Chapter 3: Motion in One Dimension PDF

E3 60 86 Motion In One Dimension 2.1 Position. ID Any object is situated at point O and three observers from three different places are looking for same object, then all three observers will have different observations about the position of point O and no one will be U wrong. Because they are observ...

E3 60 86 Motion In One Dimension 2.1 Position. ID Any object is situated at point O and three observers from three different places are looking for same object, then all three observers will have different observations about the position of point O and no one will be U wrong. Because they are observing the object from their different positions. Observer ‘A’ says : Point O is 3 m away in west direction. YG Observer ‘B’ says : Point O is 4 m away in south direction. Observer ‘C’ says : Point O is 5 m away in east direction. N W O E S B C 5m Therefore position of any point is completely expressed 4m 3m A by two factors: Its distance from the observer and its direction D with respect to observer. U That is why position is characterised by a vector known as position vector.  Let point P is in a xy plane and its coordinates are (x, y). Then position vector (r ) of point will be xˆi  yˆj ST and if the point P is in a space and its coordinates are ( x, y, z) then position vector can be expressed as  r  xˆi  yˆj  zkˆ. 2.2 Rest and Motion. If a body does not change its position as time passes with respect to frame of reference, it is said to be at rest. And if a body changes its position as time passes with respect to frame of reference, it is said to be in motion. Frame of Reference : It is a system to which a set of coordinates are attached and with reference to which observer describes any event. Motion In One Dimension 87 A passenger standing on platform observes that tree on a platform is at rest. But when the same passenger is passing away in a train through station, observes that tree is in motion. In both conditions observer is right. But observations are different because in first situation observer stands on a platform, which is reference frame 60 at rest and in second situation observer moving in train, which is reference frame in motion. So rest and motion are relative terms. It depends upon the frame of references. E3 Tree is at rest motion ID Platform (Frame of reference) Tree is in 2.3 Types of Motion. YG U Moving train (Frame of reference) Two dimensional Three dimensional Motion of a body in a straight line is Motion of body in a plane is called Motion of body in a space is called called one dimensional motion. two dimensional motion. three dimensional motion. When only one coordinate of the When two coordinates of the position When all three coordinates of the position of a body changes with time of a body changes with time then it is position of a body changes with time then it is said to be moving one said to be moving two dimensionally. then it is said to be moving three ST U D One dimensional dimensionally. e.g.. Motion of car on a straight road. Motion of freely falling body. dimensionally. e.g. Motion of car on a circular turn. Motion of billiards ball. e.g.. Motion of flying kite. Motion of flying insect. 2.4 Particle or Point Mass. The smallest part of matter with zero dimension which can be described by its mass and position is defined as a particle. 88 Motion In One Dimension If the size of a body is negligible in comparison to its range of motion then that body is known as a particle. 60 A body (Group of particles) to be known as a particle depends upon types of motion. For example in a planetary motion around the sun the different planets can be presumed to be the particles. In above consideration when we treat body as particle, all parts of the body undergo same displacement E3 and have same velocity and acceleration. 2.5 Distance and Displacement. ID (1) Distance : It is the actual path length covered by a moving particle in a given interval of time. (i) If a particle starts from A and reach to C through point B as shown in the figure. C (ii) Distance is a scalar quantity. (iv) Unit : metre (S.I.) 4m A 3m B YG (iii) Dimension : [M0L1T0] U Then distance travelled by particle  AB  BC  7 m (2) Displacement : Displacement is the change in position vector i.e., A vector joining initial to final position. D (i) Displacement is a vector quantity U (ii) Dimension : [M0L1T0] (iii) Unit : metre (S.I.) ST (iv) In the above figure the displacement of the particle AC  AB  BC  | AC |  ( AB ) 2  (BC ) 2  2( AB ) (BC ) cos 90 o = 5m     (v) If S 1 , S 2 , S 3........ S n are the displacements of a body then the total (net) displacement is the vector sum      of the individuals. S  S 1  S 2  S 3 ........  S n (3) Comparison between distance and displacement : (i) The magnitude of displacement is equal to minimum possible distance between two positions. So distance  |Displacement|. Motion In One Dimension 89 (ii) For a moving particle distance can never be negative or zero while displacement can be. (zero displacement means that body after motion has came back to initial position) 60 i.e., Distance > 0 but Displacement > = or < 0 (iii) For motion between two points displacement is single valued while distance depends on actual path E3 and so can have many values. (iv) For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position. along a straight line without change in direction. ID (v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is Y B   (vi) If rA and rB are the position vectors of particle initially and finally. A  rA X YG    rAB  rB  rA  rAB  rB U Then displacement of the particle s and s is the distance travelled if the particle has gone through the path APB. A man goes 10m towards North, then 20m towards east then displacement is U Problem 1. D Sample problems based on distance and displacement (b) 25m (c) 25.5m (d) 30m ST (a) 22.5m [KCET (Med.) 1999; JIPMER 1999; AFMC 2003] Solution : (a) If we take east as x  axis and north as y  axis, then displacement  20 ˆi  10 ˆj So, magnitude of displacement  20 2  10 2  10 5 = 22.5 m. Problem 2. A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled and displacement will be respectively (a) Solution : (a) r 2 ,r 2 (b) r 4 ,r (c) r, r (d) r, r 2  Let particle start from A, its position vector rO A  rˆi Y O B A X 90 Motion In One Dimension  After one quarter position vector rOB  r ˆj. So displacement  rˆj  rˆi and distance = one fourth of circumference  2r r  4 2 The displacement of the point of the wheel initially in contact with the ground, when the wheel roles E3 Problem 3. 60 Magnitude of displacement  r 2. forward half a revolution will be (radius of the wheel is R) (a) R (b) R  2  4  4 (c) 2R 2 Horizontal distance covered by the wheel in half revolution = R ID Solution : (b) (d) R So the displacement of the point which was initially in contact with a  4. 2.6 Speed and Velocity. 2R Pinitial R YG R  2 (R)  (2 R) 2 U ground = 2 Pnew (1) Speed : Rate of distance covered with time is called speed. D (i) It is a scalar quantity having symbol . U (ii) Dimension : [M0L1T–1] (iii) Unit : metre/second (S.I.), cm/second (C.G.S.) ST (iv) Types of speed : (a) Uniform speed : When a particle covers equal distances in equal intervals of time, (no matter how small the intervals are) then it is said to be moving with uniform speed. In given illustration motorcyclist travels equal distance (= 5m) in each second. So we can say that particle is moving with uniform speed of 5 m/s. Distance Time Uniform Speed 5m 1 sec 5m 1 sec 5m/ 5m/ s s 5m 5m 5m 1 sec 1 sec 1 sec 5m/s 5m/ 5m/ s s 5m 1m/s 5m/s Motion In One Dimension 91 60 (b) Non-uniform (variable) speed : In non-uniform speed particle covers unequal distances in equal intervals of time. In the given illustration motorcyclist travels 5 m in 1st second, 8m in 2nd second, 10m in 3rd E3 second, 4m in 4th second etc. Therefore its speed is different for every time interval of one second. This means particle is moving with ID variable speed. 5m 8m 10m 4m 6m 7m Time 1 sec 1 sec 1 sec 1 sec 1 sec 1 sec 5m/ 8m/ 10m/s 4m/ 6m/ 7m/s s s s s YG Variable Speed U Distance (c) Average speed : The average speed of a particle for a given ‘Interval of time’ is defined as the ratio of D distance travelled to the time taken. Distance travelled s ; v av  t Time taken U Average speed   Time average speed : When particle moves with different uniform speed  1 ,  2 ,  3... etc in different ST time intervals t 1 , t 2 , t 3 ,... etc respectively, its average speed over the total time of journey is given as v av  Total distance covered Total time elapsed   t   2 t 2   3 t 3 ...... d1  d 2  d 3 ...... = 11 t1  t 2  t 3 ...... t1  t2  t3 ...... Special case : When particle moves with speed v1 upto half time of its total motion and in rest time it is moving with speed v2 then v av  v1  v 2 2 92 Motion In One Dimension  Distance averaged speed : When a particle describes different distances d 1 , d 2 , d 3 ,...... with different time intervals t 1 , t 2 , t 3 ,...... with speeds v 1 , v 2 , v 3...... respectively then the speed of particle averaged over the  av  d  d 2  d 3 ...... Total distance covered d  d 2  d 3 ......  1  1 d1 d 2 d 3 Total time elapsed t1  t2  t3 ......   ...... 1 2 3 60 total distance can be given as E3  When particle moves the first half of a distance at a speed of v1 and second half of the distance at speed v2 then 2v 1 v 2 v1  v 2 ID v av   When particle covers one-third distance at speed v1, next one third at speed v2 and last one third at speed v3, then 3 v1v 2 v 3 v1v 2  v 2 v 3  v 3 v1 U v av  YG (d) Instantaneous speed : It is the speed of a particle at particular instant. When we say “speed”, it usually means instantaneous speed. The instantaneous speed is average speed for infinitesimally small time interval ( i.e., t 0 ). Thus Instantaneous speed v  lim D t 0 s ds  t dt U (2) Velocity : Rate of change of position i.e. rate of displacement with time is called velocity. (i) It is a scalar quantity having symbol v. ST (ii) Dimension : [M0L1T–1] (iii) Unit : metre/second (S.I.), cm/second (C.G.S.) (iv) Types (a) Uniform velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line without reversing its direction. Motion In One Dimension 93 (b) Non-uniform velocity : A particle is said to have non-uniform velocity, if either of magnitude or direction of velocity changes (or both changes). Instantaneous velocity is defined as rate of change of position vector of E3 (d) Instantaneous velocity : 60 (c) Average velocity : It is defined as the ratio of displacement to time taken by the body   Displaceme nt r ; v av  Average velocity  Time taken t ID particles with time at a certain instant of time.     r dr Instantaneous velocity v  lim  t 0 t dt (v) Comparison between instantaneous speed and instantaneous velocity (a) instantaneous velocity is always tangential to the path followed by Y U the particle. 3 2 YG When a stone is thrown from point O then at point of projection the instantaneous velocity of stone is v 1 , at point A the instantaneous velocity A 4 B v1 C O X of stone is v 2 , similarly at point B and C are v 3 and v 4 respectively. Direction of these velocities can be found out by drawing a tangent on the trajectory at a given point. D (b) A particle may have constant instantaneous speed but variable instantaneous velocity. U Example : When a particle is performing uniform circular motion then for every instant of its circular motion its speed remains constant but velocity changes at every instant. ST (c) The magnitude of instantaneous velocity is equal to the instantaneous speed. (d) If a particle is moving with constant velocity then its average velocity and instantaneous velocity are always equal. (e) If displacement is given as a function of time, then time derivative of displacement will give velocity.  Let displacement x  A0  A1t  A2 t 2   dx d Instantaneous velocity v   ( A0  A1 t  A 2 t 2 ) dt dt  v   A1  2 A 2 t 94 Motion In One Dimension For the given value of t, we can find out the instantaneous velocity.   e.g. for t  0 ,Instantaneous velocity v   A1 and Instantaneous speed | v |  A1 60 (vi) Comparison between average speed and average velocity (a) Average speed is scalar while average velocity is a vector both having same units ( m/s) and E3 dimensions [LT 1 ]. (b) Average speed or velocity depends on time interval over which it is defined. (c) For a given time interval average velocity is single valued while average speed can have many values ID depending on path followed.     (d) If after motion body comes back to its initial position then v av  0 (as r  0 ) but v av  0 and finite U as (s  0). YG (e) For a moving body average speed can never be negative or zero (unless t ) while average velocity  can be i.e. v av  0 while v a = or < 0. Sample problems based on speed and velocity If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2 then average speed is 1 v 1v 2 2 (b) v1  v 2 2 U (a) D Problem 4. Average speed = x Total distance travelled  t1  t 2 Total time taken ST Solution : (d) 5v 1 v 2  = (2 / 5) x (3 / 5)x 2v 2  3v 1  v1 v2 x Problem 5. Solution : (a) [MP PMT 2003] (c) 2v 1v 2 v1  v 2 (d) 5v 1v 2 3v 1  2v 2 (2/5)x (3/5)x t1 t2 A car accelerated from initial position and then returned at initial point, then (a) Velocity is zero but speed increases (b) Speed is zero but velocity increases (c) Both speed and velocity increase (d) Both speed and velocity decrease As the net displacement = 0 Hence velocity = 0 ; but speed increases. [AIEEE 2002] Motion In One Dimension 95 Note Problem 6. | Average velocity |  1  | Av.speed |  | Av. velocity | | Average speed | : A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the man over the interval of time 0 to 40 min. is equal to Solution : (d) (b) 25 km/h 4 Time taken in going to market  (c) 30 km/h 4 (d) 45 km/h 8 E3 (a) 5 km/h 60 the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of 2.5 1  hr  30 min. 5 2 ID As we are told to find average speed for the interval 40 min., so remaining time for consideration of motion is 10 min. Problem 7. (2.5  1.25 ) km 45 Total distance  km / hr. = (40 / 60 ) hr. 8 Total time YG Hence, average speed = 10  1.25 km. 60 U So distance travelled in remaining 10 min = 7.5  The relation 3 t  3 x  6 describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is (c) 5 metres (d) Zero  3 x  (3t  6)  3 x  (3 t  6) 2  x  3t 2  12t  12 3t  3x  6 v= dx d  (3 t 2  12 t  12 )  6 t  12 dt dt U Solution : (d) (b) 12 metres D (a) 24 metres ST If velocity = 0 then, 6 t  12  0  t  2 sec Hence at t = 2, x = 3(2)2 – 12 (2) + 12 = 0 metres. Problem 8. The motion of a particle is described by the equation x  a  bt 2 where a  15 cm and b  3 cm. Its instantaneous velocity at time 3 sec will be (a) 36 cm/sec Solution : (b) x  a  bt 2 (b) 18 cm/sec v= [AMU (Med.) 2000] (c) 16 cm/sec dx  0  2bt dt At t = 3sec, v = 2  3  3 = 18 cm / sec (As b  3cm ) (d) 32 cm/sec 96 Motion In One Dimension Problem 9. A train has a speed of 60 km/h for the first one hour and 40 km/h for the next half hour. Its average speed in km/h is [JIPMER 1999] Total distance travelled = 60  1  40   Average speed  Problem 10. (c) 48 1 1 3  80 km and Total time taken = 1 hr  hr  hr 2 2 2 80  53.33 km/h 3 2 A person completes half of its his journey with speed 1 and rest half with speed 2. The average speed of the person is 1   2 (b)   2 21  2 1   2 (c)   1  2 1   2 (d)   1  2 In this problem total distance is divided into two equal parts. So d d  d1  d 2 2 2  = d1 d 2 d/2 d/2   2 1 2  av  2 1 1  1 2 YG 1 2 1  2 1   2 U av Problem 11. [RPET 1993; MP PMT 2001] ID (a)   Solution : (b) (d) 70 E3 Solution : (b) (b) 53.33 60 (a) 50 = A car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr. The average speed is Solution : (d) (b) 80 km/hr D (a) 40 km/hr [MP PMT 1999; CPMT 2002] (c) 46 Let total distance travelled = x and total time taken t1 + t2  U  Average speed  2 km /hr 3 (d) 36 km/hr x / 3 2x / 3  20 60 x 1 = 36 km / hr  (1 / 3)x (2 / 3)x 1 2   20 60 60 180 ST 2.7 Acceleration. The time rate of change of velocity of an object is called acceleration of the object. (1) It is a vector quantity. It’s direction is same as that of change in velocity (Not of the velocity) (2) There are three possible ways by which change in velocity may occur When only direction of velocity When only magnitude of velocity When both magnitude and direction changes changes of velocity changes Acceleration parallel or anti-parallel Acceleration has two components one Acceleration perpendicular to Motion In One Dimension 97 velocity is to velocity perpendicular to velocity and another parallel or anti-parallel to e.g. Uniform circular motion 60 velocity e.g. Motion under gravity e.g. Projectile motion (4) Unit : metre/second2 (S.I.); cm/second2 (C.G.S.) (5) Types of acceleration : E3 (3) Dimension : [M0L1T–2] ID (i) Uniform acceleration : A body is said to have uniform acceleration if magnitude and direction of the acceleration remains constant during particle motion. If a particle is moving with uniform acceleration, this does not necessarily imply that particle is U Note :  moving in straight line. e.g. Projectile motion. YG (ii) Non-uniform acceleration : A body is said to have non-uniform acceleration, if magnitude or direction or both, change during motion.     v v 2  v 1 (iii) Average acceleration : a a    t t U D   v The direction of average acceleration vector is the direction of the change in velocity vector as a  t    v dv (iv) Instantaneous acceleration = a  lim  t 0 t dt ST (v) For a moving body there is no relation between the direction of instantaneous velocity and direction of acceleration.  a  a Y  a  g  2 1 O  g g  3 X 98 Motion In One Dimension e.g. (a) In uniform circular motion  = 90º always 60 (b) In a projectile motion  is variable for every point of trajectory. E3   F (vi) If a force F acts on a particle of mass m, by Newton’s 2nd law, acceleration a  m     dv d 2 x   dx  (vii) By definition a   2  As v  dt  dt dt  i.e., if x is given as a function of time, second time derivative of displacement gives acceleration dv dv dx d    v. dt dx dt dx ID (viii) If velocity is given as a function of position, then by chain rule a  dx   as v  dt    (ix) If a particle is accelerated for a time t1 by acceleration a1 and for time t2 by acceleration a2 then average a1t1  a2 t2 t1  t2 U acceleration is aa  YG (x) If same force is applied on two bodies of different masses m 1 and m 2 separately then it produces accelerations a1 and a 2 respectively. Now these bodies are attached together and form a a1 F a2 F m1 D combined system and same force is applied on that system so that a be the acceleration of the combined system, then F m1 m2 F F F   a a1 a 2 aa 1 1 1  a 1 2   a1  a 2 a a1 a 2 ST So, a U F  m 1  m 2  a  m2 (xi) Acceleration can be positive, zero or negative. Positive acceleration means velocity increasing with time, zero acceleration means velocity is uniform constant while negative acceleration (retardation) means velocity is decreasing with time. (xii) For motion of a body under gravity, acceleration will be equal to “g”, where g is the acceleration due to gravity. Its normal value is 9.8 m/s2 or 980 cm/s2 or 32 feet/s2. Sample problems based on acceleration Motion In One Dimension 99 Problem 12. The displacement of a particle, moving in a straight line, is given by s  2t 2  2t  4 where s is in metres and t in seconds. The acceleration of the particle is (b) 4 m/s2 (c) 6 m/s2 (d) 8 m/s2 dv ds  4 (1)  0  4 m / s 2  4 t  2 and acceleration (a)  dt dt 60 (a) 2 m/s2 [CPMT 2001] Solution : (b) Given s  2t 2  2t  4  velocity (v)  Problem 13. The position x of a particle varies with time t as x  at 2  bt 3. The acceleration of the particle will be zero at time t equal to a b (b) E3 (a) [CBSE PMT 1997; BHU 1999; DPMT 2000; KCET (Med.) 2000] 2a 3b (c) a 3b (d) Zero dv dx  2a  6 bt.  2at  3bt 2 and acceleration (a) = dt dt 2a a When acceleration = 0  2a  6bt  0  t .  6 b 3b Given x  at 2  bt 3  velocity (v)  Problem 14. The displacement of the particle is given by y  a  bt  ct 2  dt 4. The initial velocity and acceleration are ID Solution : (c) U respectively (a) b,4 d Given y  a  bt  ct 2  dt 4  v = (c) b, 2 c (d) 2c,  4 d dy  0  b  2ct  4 dt 3 dt YG Solution : (c) (b) b, 2 c [CPMT 1999, 2003] Putting t  0, vinitial = b So initial velocity = b dv  0  2c  12 d t 2 dt D Now, acceleration (a)  Putting t = 0, ainitial = 2c The relation between time t and distance x is t   x 2   x, where  and  are constants. The retardation U Problem 15. is (v is the velocity) ST (a) 2v 3 Solution : (a) (b) 2 v 3 differentiating time with respect to distance So, acceleration (a) = Problem 16. [NCERT 1982] (c) 2v 3 (d) 2  2 v 3 dx 1 dt   2x    v  dt 2x   dx v. 2 dv dv dv dx   2.v.v 2  2v 3 . =v 2 dx (2x   ) dt dx dt If displacement of a particle is directly proportional to the square of time. Then particle is moving with [RPET 1999] (a) Uniform acceleration (c) Uniform velocity Solution : (a) Given that x  t 2 or x  Kt 2 (where K= constant) (b) Variable acceleration (d) Variable acceleration but uniform velocity 100 Motion In One Dimension dx d  2 Kt and Acceleration (a)   2K dt dt Velocity (v)  It is clear that velocity is time dependent and acceleration does not depend on time. Problem 17. 60 So we can say that particle is moving with uniform acceleration but variable velocity. A particle is moving eastwards with velocity of 5 m/s. In 10 sec the velocity changes to 5 m/s northwards. The average acceleration in this time is [IIT-JEE 1982] (a) Zero (b) 1 1 (c) Solution : (b) 2   m/s 2 toward north-east (d)    2  1       212 cos 90  5  5  5 2 2 2 o 2   5 2  5 2 1   m/s 2 toward northt 10 2 Problem 18.  v1    2  5m / s 90o  1  5 m / s YG west (As clear from the figure). U Average acceleration  2 1 m/s 2 toward north-west 2 ID 2 1 m/s 2 toward north-west E3 2 A body starts from the origin and moves along the x-axis such that velocity at any instant is given by D (4 t 3  2 t) , where t is in second and velocity is in m/s. What is the acceleration of the particle, when it is 2 m from the origin? (b) 22 m/s 2 U (a) 28 m/s 2 (c) 12 m /s 2 Given that   4 t 3  2t  x   dt , x  t 4  t 2  C , at t  0, x  0  C  0 ST Solution : (b) [IIT-JEE 1982] When particle is 2m away from the origin 2  t 4  t 2  t 4  t 2  2  0  (t 2  2) (t 2  1)  0  t  a   d d  4 t 3  2 t  12 t 2  2  a  12t 2  2 dt dt for t  2 sec  a  12   2 2  2  a  22 m /s 2 2 sec (d) 10 m /s 2 Motion In One Dimension 101 Problem 19. A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 sec on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is Solution : (b) (b)  3 m/s 2 (c) 0.3 m/s 2 (d)  0.3 m/s 2 60 (a) 3 m/s 2 [MP PET 1997] Let particle moves towards east and by the application of constant force it moves towards west    a Change in velocity  2  1  Time t E3 1  10 m/s and  2  2 m/s. Acceleration  (2)  (10 ) 12   3 m /s 2 4 4 ID 2.8 Position Time Graph. During motion of the particle its parameters of kinematical analysis ( u, v, a, r) changes with time. This can U be represented on the graph. Position time graph is plotted by taking time t along x-axis and position of the particle on y-axis. y  y1 Change in position  2 Time taken t 2  t1 D BC AD y 2  y 1 From triangle ABC tan     AC AC t 2  t1 U By comparing (i) and (ii) …(i) ….(ii) y Position As Velocity = YG Let AB is a position-time graph for any moving particle y1 O Velocity = tan v = tan ST It is clear that slope of position-time graph represents the velocity of the particle. Various position – time graphs and their interpretation D y2 B  A t1 C t2 Time x 102 Motion In One Dimension P  = 0o so v = 0 T i.e., line parallel to time axis represents that the particle is at rest. 60 O P E3  = 90o so v =  i.e., line perpendicular to time axis represents that particle is changing its position O T but time does not changes it means the particle possesses infinite velocity. Practically this is not possible. ID P  = constant so v = constant, a = 0 O T U i.e., line with constant slope represents uniform velocity of the particle. P P O O i.e., line bending towards position axis represents increasing velocity of particle. It means the particle possesses acceleration. D T T YG  is increasing so v is increasing, a is positive. ST U  is decreasing so v is decreasing, a is negative i.e., line bending towards time axis represents decreasing velocity of the particle. It means the particle possesses retardation. P O  T  constant but > 90o so v will be constant but negative i.e., line with negative slope represent that particle returns towards the point of reference. (negative displacement). Motion In One Dimension 103 P B O Straight line segments of different slopes represent that velocity of the body C S T changes after certain interval of time. T This graph shows that at one instant the particle has two positions. Which is not 60 A O E3 P possible. ID P The graph shows that particle coming towards origin initially and after that it is moving away from origin. T Note :  If YG U O the graph is plotted between distance and time then it is always an increasing curve and it time. Hence such type of distance time graph is valid up to point A D only, after point A it is not valid as shown in the figure. ST U  For two particles having displacement time graph with slopes 1 and 2 tan 1  possesses velocities v1 and v2 respectively then 1   2 tan  2 Problem 20. Distance never comes back towards origin because distance never decrease with O A Time Sample problems based on position-time graph The position of a particle moving along the x-axis at certain times is given below : t (s) 0 1 2 3 x (m) –2 0 6 16 Which of the following describes the motion correctly (a) Uniform, accelerated (b) (c) Non-uniform, accelerated Solution : (a) Instantaneous velocity v  Uniform, decelerated (d) There is not enough data for generalisation x , By using the data from the table t 104 Motion In One Dimension 0  (2) 6 0 16  6  2m /s , v 2   6 m /s and v 3   10 m /s i.e. the speed is increasing at a constant 1 1 1 rate so motion is uniformly accelerated. v1  Which of the following graph represents uniform motion (b) (c) s s 60 (a) [DCE 1999] s s t t (d) E3 Problem 21. t t When distance time graph is a straight line with constant slope than motion is uniform. Problem 22. The displacement-time graph for two particles A and B are straight lines inclined at angles of 30o and 60o ID Solution : (a) [CPMT 1990; MP PET 1999; MP PET 2001] vA tan 30 o 1/ 3    o vB tan 60 3  with the time axis. The ratio of velocities of vA : vB is (b) 1 : 3 (c) U (a) 1 : 2 (d) 1 : 3 3 :1 1 1 3 Solution : (d) v  tan  from displacement graph. So Problem 23. From the following displacement time graph find out the velocity of a moving body 3 m/s In first instant you will apply   tan  and say,   tan 30 o  ST Solution : (c) O 30o Displacement (meter) 1 3 U (d) D (b) 3 m/s Time (sec) m/s 3 (c) YG 1 (a) 3 3 1 m/s. 3 But it is wrong because formula   tan  is valid when angle is measured with time axis. Here angle is taken from displacement axis. So angle from time axis  90 o  30 o  60 o. Now   tan 60 o  3 Problem 24. The diagram shows the displacement-time graph for a particle moving in a straight line. The average velocity for the interval t = 0, t = 5 is x 20 10 (a) 0 O (b) 6 ms –1 – 10 5 2 4 t Motion In One Dimension 105 (c) – 2 ms–1 (d) 2 ms–1 Total displaceme nt 20   20   10  = –2 m/s = Total time 5 Average velocity = Problem 25. Figure shows the displacement time graph of a body. What is the ratio of the speed in the first second and 60 Solution : (c) Y (a) 1 : 2 (b) 1 : 3 20 10 0 1 2 3 Time X ID (c) 3 : 1 (d) 2 : 1 Speed in first second = 30 and Speed in next two seconds = 15. So that ratio 2 : 1 2.9 Velocity Time Graph. U Solution: (d) 30 E3 Displacement that in the next two seconds YG The graph is plotted by taking time t along x-axis and velocity of the particle on y-axis. Distance and displacement : The area covered between the velocity time graph and time axis gives the displacement and distance travelled by the body for a given time interval. + D Then Total distance | A1 | | A 2 | | A 3 | 1  = Addition of modulus of different area. i.e. s  |  | dt t 2 Total displacement  A1  A 2  A 3 U 3 –  ST = Addition of different area considering their sign. i.e. r   dt here A1 and A2 are area of triangle 1 and 2 respectively and A3 is the area of trapezium. Acceleration : Let AB is a velocity-time graph for any moving particle Change in velocity v 2  v 1  Time taken t 2  t1 BC AD v 2  v1 From triangle ABC, tan     AC AC t 2  t1 By comparing (i) and (ii) Acceleration (a) = tan  Velocity As Acceleration = y …(i) D v2 B  v1 A ….(ii) O t1 Time C t2 x 106 Motion In One Dimension ST U D YG U ID E3 60 It is clear that slope of velocity-time graph represents the acceleration of the particle. Motion In One Dimension 107 Velocity Various velocity – time graphs and their interpretation  = 0, a = 0, v = constant 60 i.e., line parallel to time axis represents that the particle is moving with constant Time velocity.  = 90o, a = , v = increasing i.e., line perpendicular to time axis represents that the particle is increasing its velocity, Time ID but time does not change. It means the particle possesses infinite acceleration. Practically it is not possible. Velocity O Time U O E3 Velocity O  =constant, so a = constant and v is increasing uniformly with time O YG Velocity i.e., line with constant slope represents uniform acceleration of the particle.  increasing so acceleration increasing i.e., line bending towards velocity axis represent the increasing acceleration in the Time Time U O D Velocity body.  decreasing so acceleration decreasing Velocity ST i.e. line bending towards time axis represents the decreasing acceleration in the body O Time Positive constant acceleration because  is constant and < 90o but initial velocity of the particle is negative. Velocity 108 Motion In One Dimension Negative constant acceleration because  is constant and > 90o but initial velocity of the particle is zero. Time U O ID Negative constant acceleration because  is constant and > 90o but initial velocity of Velocity O E3 the particle is positive. Time Velocity O 60 particle is positive. Time Velocity O Positive constant acceleration because  is constant and < 90o but initial velocity of Time YG Negative constant acceleration because  is constant and > 90o but initial velocity of the particle is negative. Sample problems based on velocity-time graph A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the D Problem 26. Solution : (c) Time (c) Time (d) Speed (b) Speed (a) Speed ST Speed U ball during its flight if the air resistance is not ignored Time Time In first half of motion the acceleration is uniform & velocity gradually decreases, so slope will be negative but for next half acceleration is positive. So slope will be positive. Thus graph ' C' is correct. Not ignoring air resistance means upward motion will have acceleration (a + g) and the downward motion will have (g  a). Motion In One Dimension 109 Problem 27. A train moves from one station to another in 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is [Kerala (Engg.) 2002] 60 80 D 60 40 20 A B C N M L 0.25 0.75 1.00 E3 Speed in km/hours 100 E 1.5 2.00 (c) 100 km h–2 that slope is for line CD. So, a max  slope of CD = 60  20 40   160 km h  2. 1.25  1.00 0.25 The graph of displacement v/s time is S D Problem 28. (d) 120 km h–2 Maximum acceleration means maximum slope in speed – time graph. YG Solution : (b) (b) 160 km h–2 U (a) 140 km h–2 ID Time in hours ST U t Its corresponding velocity-time graph will be (a) (b) V (c) V t [DCE 2001] (d) V V t t t 110 Motion In One Dimension Solution : (a) We know that the velocity of body is given by the slope of displacement – time graph. So it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of the graph) and then it will be negative. In the following graph, distance travelled by the body in metres is 60 Problem 29. [EAMCET 1994] Y (b) 250 Distance = The area under v – t graph For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its 1 2 1 4 (c) 1 3 0 3 5 Time (s) D 7 2 3 ST Distance covered in total 7 seconds = Area of trapezium ABCD  Distance covered in last 2 second = area of triangle CDQ= So required fraction  1 (2  6)  10 = 40 m 2 1  2  10  10 m 2 10 1  40 4 The velocity time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 sec are respectively (a) 8 m , 16 m (b) 16 m , 8 m 4 v (m/s) Problem 31. Q P 1 U (d) C 5 A D (b) [MP PMT/PET 1998; RPET 2001] B 10 v (m/s ) (a) YG motion is what fraction of the total distance covered by it in all the seven seconds Solution : (b) X 30 Time (s) U Problem 30. 1 (30  10)  10 = 200 metre 2 20 10 ID (d) 400 S  5 0 (c) 300 Solution : (a) 10 E3 v (m/s ) (a) 200 15 A B G 2 2 O –2 –4 6 4 C D H F E I t (sec) Motion In One Dimension 111 (c) 16 m, 16 m (d) 8 m, 8 m Area of rectangle ABCO = 4  2 = 8 m 60 Solution : (a) Area of rectangle CDEF = 2 × (– 2) = – 4 m Area of rectangle FGHI = 2×2 = 4 m E3 Displacement = sum of area with their sign = 8 + (– 4) + 4 = 8 m Distance = sum of area with out sign = 8 + 4 + 4 = 16 m A ball is thrown vertically upward which of the following graph represents velocity time graph of the ball ID U Velocity Time (c) Time [CPMT 1993; AMU (Engg.) 2000] (d) Time Time YG Solution : (d) (b) Velocity (a) Velocity during its flight (air resistance is neglected) Velocity Problem 32. In the positive region the velocity decreases linearly (during rise) and in negative region velocity increase D linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in the negative region. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically U Problem 33. ST to a height d. Neglecting subsequent motion and air resistance, its velocity  varies with the height h 2 above the ground as. [IIT-JEE (Screening) 2000]  (a) d/2 Solution : (a)   d (b) h (c) d/2 d h When ball is dropped from height d its velocity will be zero.  (d) d/2 d h d/2 d h 112 Motion In One Dimension As ball comes downward h decreases and  increases just before the rebound from the earth h = 0 and v = maximum and just after rebound velocity reduces to half and direction becomes opposite. This interpretation is clearly shown by graph (a). The acceleration-time graph of a body is shown below – E3 Problem 34. d. 2 60 As soon as the height increases its velocity decreases and becomes zero at h  ID a U t  (a) YG The most probable velocity-time graph of the body is  (b) (d) t t t From given a  t graph acceleration is increasing at constant rate  da  k (constant)  a  kt (by integration) dt  dv  kt  dv  ktdt  dt ST U Solution : (c) (c) D t    dv  k  tdt  v  kt 2 2 i.e., v is dependent on time parabolically and parabola is symmetric about v-axis. and suddenly acceleration becomes zero. i.e. velocity becomes constant. Hence (c) is most probable graph. Problem 35. Which of the following velocity time graphs is not possible v v v (a) (b) O t v (c) O t (d) O t O t 60 Motion In One Dimension 113 Solution : (d) Particle can not possess two velocities at a single instant so graph (d) is not possible. Problem 36. For a certain body, the velocity-time graph is shown in the figure. The ratio of applied forces for intervals 1 2 (c)  1 3 (d)  1 3 A D 60o 30o t C B Ratio of applied force = Ratio of acceleration = Problem 37. ID (b)   U 1 2 YG Solution : (d) (a)  E3 AB and BC is aAB tan 30 1 3 = 1 3   aBC tan 120   3 Velocity-time graphs of two cars which start from rest at the same time, are shown in the figure. Graph Velocity D shows, that B A U A (a) Initial velocity of A is greater than the initial velocity of B B t O Time ST (b) Acceleration in A is increasing at lesser rate than in B (c) Acceleration in A is greater than in B (d) Acceleration in B is greater than in A Solution : (c) At a certain instant t slope of A is greater than B (  A   B ), so acceleration in A is greater than B Problem 38. Which one of the following graphs represent the velocity of a steel ball which fall from a height on to a Time (d) Time Velocity (c) Time Velocity (b) Velocity (a) Velocity marble floor? (Here  represents the velocity of the particle and t the time) Time 114 Motion In One Dimension Solution : (a) Initially when ball falls from a height its velocity is zero and goes on increasing when it comes down. Just 60 after rebound from the earth its velocity decreases in magnitude and its direction gets reversed. This process is repeated untill ball comes to at rest. This interpretation is well explained in graph (a). E3 The adjoining curve represents the velocity-time graph of a particle, its acceleration values along OA, AB and BC in metre/sec2 are respectively ID (a) 1, 0, – 0.5 (b) 1, 0, 0.5 (d) 1, 0.5, 0 A B 5 O 10 20 30 40 Time (sec) v 2  v 1 10  0   1m / s 2 t 10 0 Acceleration along OB  0 10 0  10 Acceleration along BC   – 0.5 m/s2 20 Acceleration along OA  YG Solution : (a) 10 U (c) 1, 1, 0.5 Velocity (m/sec) Problem 39. D 2.10 Equations of Kinematics. as : U These are the various relations between u, v, a, t and s for the moving particle where the notations are used ST u = Initial velocity of the particle at time t = 0 sec v = Final velocity at time t sec a = Acceleration of the particle s = Distance travelled in time t sec sn = Distance travelled by the body in nth sec (1) When particle moves with zero acceleration (i) It is a unidirectional motion with constant speed. Motion In One Dimension 115 (ii) Magnitude of displacement is always equal to the distance travelled. (iii) v = u, s=ut [As a = 0] 60 (2) When particle moves with constant acceleration (i) Acceleration is said to be constant when both the magnitude and direction of acceleration remain E3 constant. (ii) There will be one dimensional motion if initial velocity and acceleration are parallel or anti-parallel to each other. Equation of motion in vector from ID (iii) Equations of motion in scalar from    v  u  at s  ut  1 2 at 2 YG  2  u 2  2as U   u  at u v  s t  2  sn  u  a (2n  1) 2   1 s  u t  at 2 2    v.v  u.u  2a.s  1   s  (u  v ) t 2    a s n  u  (2n  1) 2 D (3) Important points for uniformly accelerated motion U (i) If a body starts from rest and moves with uniform acceleration then distance covered by the body in t sec is proportional to t2 (i.e. s  t 2 ). ST So we can say that the ratio of distance covered in 1 sec, 2 sec and 3 sec is 12 : 2 2 : 3 2 or 1 : 4 : 9. (ii) If a body starts from rest and moves with uniform acceleration then distance covered by the body in nth sec is proportional to (2n  1) (i.e. s n  (2n  1) So we can say that the ratio of distance covered in I sec, II sec and III sec is 1 : 3 : 5. (iii) A body moving with a velocity u is stopped by application of brakes after covering a distance s. If the same body moves with velocity nu and same braking force is applied on it then it will come to rest after covering a distance of n2s. As  2  u 2  2 as  0  u 2  2as  s  u2 , s  u 2 [since a is constant] 2a 116 Motion In One Dimension So we can say that if u becomes n times then s becomes n2 times that of previous value. (iv) A particle moving with uniform acceleration from A to B along a straight line has velocities  1 and  2 60 at A and B respectively. If C is the mid-point between A and B then velocity of the particle at C is equal to  12   22  2 Problem 40. E3 Sample problems based on uniform acceleration A body A moves with a uniform acceleration a and zero initial velocity. Another body B, starts from the The value of t is (a) v a (c) Let they meet after time 't'. Distance covered by body A = 1 2 2v. at  vt  t  2 a v 2a (d) [MP PET 2003] v 2a 1 2 at ; Distance covered by body B = vt 2 YG and Problem 41. (b) U Solution : (a) 2v a ID same point moves in the same direction with a constant velocity v. The two bodies meet after a time t. A student is standing at a distance of 50metres from the bus. As soon as the bus starts its motion with an acceleration of 1ms–2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u , so that the students is able to catch the bus D is (b) 8 ms–1 U (a) 5 ms–1 Solution : (c) [KCET 2003] (c) 10 ms–1 (d) 12 ms–1 Let student will catch the bus after t sec. So it will cover distance ut. ST Similarly distance travelled by the bus will be ut  50  1 2 at for the given condition 2 50 t 1 2 t2  u  (As a  1 m / s 2 ) at  50  t 2 2 2 To find the minimum value of u, du  0 , so we get t = 10 sec dt then u = 10 m/s. Problem 42. A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is (a) 6m (b) 12m (c) 18m (d) 24m Motion In One Dimension 117 Solution : (d) v 2  u 2  2as  0  u 2  2as  s  2 u2  s  u 2 (As a = constant) 2a s2  u2   100       s 2  4 s1  4  12  24 m. s1  u1   50  The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be u  200 m / s , v  100 m / s , s  0.1 m a u 2  v 2 (200 )2  (100 )2   15  10 4 m / s 2 2s 2  0.1 (c) 13.5  104 m/s2 (d) 15  104 m/s2 A body A starts from rest with an acceleration a1. After 2 seconds, another body B starts from rest with an ID Problem 44. (b) 12  104 m/s2 [AIIMS 2001] E3 (a) 10  104 m/s2 Solution : (d) 60 Problem 43. 2 acceleration a 2. If they travel equal distances in the 5th second, after the start of A, then the ratio a1 : a 2 is equal to (b) 5 : 7 (c) 9 : 5 U (a) 5 : 9 [AIIMS 2001] (d) 9 : 7 a 2n  1 , Distance travelled by body A in 5th second = 0  a1 (2  5  1) 2 2 a Distance travelled by body B in 3rd second is = 0  2 (2  3  1) 2 a 5 a a According to problem : 0  1 (2  5  1) = 0  2 (2  3  1)  9 a1  5 a2  1  a2 9 2 2 By using S n  u  Problem 45. The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms–1. D YG Solution : (a) If the change in velocity of the body is 0.18ms–1 during this time, its uniform acceleration is U (a) 0.01 ms–2 Solution : (b) Time  (c) 0.03 ms–2 (d) 0.04 ms–2 Distance 3.06   9 sec Average velocity 0.34 ST and Acceleration  Problem 46. (b) 0.02 ms–2 [EAMCET (Med.) 2000] Change in velocity 0.18   0.02 m / s 2. 9 Time A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec (a) 8.3 m Solution : (a) (b) 9.3 m (c) 10.3 m Let initial (t  0) velocity of particle = u for first 5 sec of motion s5  10 metre , so by using s  ut  10  5u  1 a(5)2  2u  5 a  4 2 for first 8 sec of motion s8  20 metre …. (i) 1 2 at 2 (d) None of above 118 Motion In One Dimension 1.... (ii) a(8 )2  2u  8 a  5 2 7 1 By solving (i) and (ii) u  m / s a  m / s 2 6 3 20  8 u  1 a10 2 2 60 Now distance travelled by particle in total 10 sec. s10  u  10  by substituting the value of u and a we will get s10  28.3 m So the distance in last 2 sec = s10 – s8  28.3  20  8.3 m A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S 1 , S 2 and S 3 E3 Problem 47. in the first five seconds, second five seconds and next five seconds respectively the relation between S 1 , S 2 and S 3 is Solution : (c) (b) 5 S 1  3 S 2  S 3 (c) S 1  1 1 S2  S3 3 5 ID (a) S 1  S 2  S 3 (d) S 1  [AMU (Engg.) 2000] 1 1 S2  S3 5 3 Since the body starts from rest. Therefore u  0. 1 25 a a(5) 2  2 2 100 a 1 100 a a  S2   S 1 = 75 S 1  S 2  a(10 ) 2  2 2 2 2 225 a 125 a 1 225 a  S3   S 2  S1 = S 1  S 2  S 3  a(15 ) 2  2 2 2 2 1 1 Thus Clearly S 1  S 2  S 3 3 5 Problem 48. YG U S1  If a body having initial velocity zero is moving with uniform acceleration 8 m / sec 2 , the distance travelled D by it in fifth second will be (a) 36 metres (b) 40 metres (c) 100 metres (d) Zero 1 1 a(2n  1)  0  (8 ) [2  5  1]  36 metres 2 2 Sn  u  Problem 49. The engine of a car produces acceleration 4m/sec2 in the car, if this car pulls another car of same mass, U Solution : (a) ST what will be the acceleration produced (a) 8 m/s2 (b) 2 m/s2 [RPET 1996] (c) 4 m/s2 (d) 1 m / s2 2 1 if F = constant. Since the force is same and the effective mass of system becomes double m a2 m a m  1  , a 2  1  2 m/s 2 a1 m2 2m 2 Solution : (b) F = ma a  Problem 50. A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second. [CBSE PMT 1993] (a) 7/5 (b) 5/7 (c) 7/3 (d) 3/7 Motion In One Dimension 119 Solution : (a) As S n  (2n  1) , S4 7  S3 5 2.11 Motion of Body Under Gravity (Free Fall). 60 The force of attraction of earth on bodies, is called force of gravity. Acceleration produced in the body by the force of gravity, is called acceleration due to gravity. It is represented by the symbol g. E3 In the absence of air resistance, it is found that all bodies (irrespective of the size, weight or composition) fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude (h t1 Let u is the initial velocity of body then time of ascent t1  u ga and h  u2 2(g  a) U where g is acceleration due to gravity and a is retardation by air resistance and for upward motion both will work vertically downward. YG For downward motion a and g will work in opposite direction because a always work in direction opposite to motion and g always work vertically downward. So 1 (g  a) t 22 2  u2 1  (g  a) t 22  2(g  a) 2 t2  u (g  a)(g  a) D h U Comparing t1 and t2 we can say that t2 > t1 since (g + a ) > (g – a) (10) A particle is dropped vertically from rest from a height. The time taken u=0 1m t1  1 1m t2  2 1 1m t3  3 2 t4  4  3 1m ST by it to fall through successive distance of 1m each will then be in the ratio of the difference in the square roots of the integers i.e. 1 , ( 2  1 ), ( 3  2 ).......( 4  3 ),......... Sample problems based on motion under gravity Problem 51. If a body is thrown up with the velocity of 15 m/s then maximum height attained by the body is (g = 10 m/s2) Motion In One Dimension 123 [MP PMT 2003] (a) 11.25 m (b) 16.2 m 2 (c) 24.5 m (d) 7.62 m 2 u (15 )   11. 25 m 2 g 2  10 H max  Problem 52. A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its motion is (g  10 m / s 2 ) 60 Solution : (a) [MP PET 2003] (b) 45m (c) 90m g 2n  1  h5 th  10 2  5  1  45 m. 2 2 (d) 125m E3 (a) 25m Solution : (b) hn  Problem 53. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its (a) (b) ut  1 2 gt 2 If ball is thrown with velocity u, then time of flight  (c) (u  gt )t u g U Solution : (a) 1 2 gt 2 ID ascent is u  u  velocity after   t  sec : v  u  g   t  = gt. g  g  (d) ut h YG So, distance in last 't' sec : 0 2  (gt) 2  2(g)h. 1  h  gt 2. 2 Problem 54. [CBSE 2003] t sec u  g   t  sec  A man throws balls with the same speed vertically upwards one after the other at an interval of 2 D seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given g  9. 8 m / s 2 ) (b) Any speed less than 19.6 m/s U (a) At least 0.8 m/s (c) Only with speed 19.6 m/s Interval of ball throw = 2 sec. ST Solution : (d) (d) More than 19.6 m/s If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than 4 sec. i.e. T  4 sec or 2U  4 sec  u  19.6 m /s g It is clear that for u  19.6 First ball will just strike the ground (in sky), second ball will be at highest point (in sky), and third ball will be at point of projection or on ground (not in sky). Problem 55. A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower (a) 100 meters (b) 320 meters [CPMT 2003] (c) 80 meters (d) 240 meters 124 Motion In One Dimension Let both balls meet at point P after time t. 1 2 gt 2 1 The distance travelled by ball B (h2 )  ut  gt 2 2 The distance travelled by ball A ( h1 )  400 m.....(ii) By adding (i) and (ii) h1  h2  ut = 400 (Given h  h1  h2  400. )  t  400 / 50  8 sec and h1  320 m, h2  80 m P h1 h2 B A very large number of balls are thrown vertically upwards in quick succession in such a way that the next E3 Problem 56. A.....(i) 60 Solution : (c) ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of ball thrown per minute is (take g  10ms 2 ) Solution : (c) (b) 80 (c) 60 (d) 40 ID (a) 120 [KCET (Med.) 2002] Maximum height of ball = 5m, So velocity of projection  u  2 gh  2  10  5 10 m / s u 1  1sec  min. g 60 U time interval between two balls (time of ascent) = So no. of ball thrown per min = 60 A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then YG Problem 57. maximum height attained by it is (Take g  10 m/s2) (a) 8 m Solution : (b) (b) 10 m [CBSE PMT 2001] (c) 12 m Let particle thrown with velocity u and its maximum height is H then H  (d) 16 m u2 2g D When particle is at a height H / 2 , then its speed is 10 m / s U u2 H From equation v 2  u 2  2 gh , 10 2  u 2  2 g    u 2  2 g  u 2  200 2 4 g   ST  Maximum height H  Problem 58. u2 200  = 10m 2 g 2  10 A stone is shot straight upward with a speed of 20 m/sec from a tower 200 m high. The speed with which it strikes the ground is approximately (a) 60 m/sec Solution : (b) (b) 65 m/sec [AMU (Engg.) 1999] (c) 70 m/sec (d) 75 m/sec Speed of stone in a vertically upward direction is 20 m/s. So for vertical downward motion we will consider u  20 m / s v 2  u 2  2 gh  (20)2  2  10  200  v  65 m / s Problem 59. A body freely falling from the rest has a velocity ‘v’ after it falls through a height ‘h’. The distance it has to fall down for its velocity to become double, is [BHU 1999] Motion In One Dimension 125 (a) 2h (c) 6h Let at point A initial velocity of body is equal to zero For path AB : v 2  0  2 gh u=0 A … (i) For path AC : (2v)  0  2 gx  4 v  2 gx 2 h … (ii) 2 x B v C 2v E3 Solving (i) and (ii) x  4 h Problem 60. (d) 8h 60 Solution : (b) (b) 4h A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top S  1 2 at  t  2 t2  t1 Problem 61. s2  s1 (c) 4 s s  As a  constant  (d) 16 s t s/4 1 4   t2  1   2 s s 2 2 2 U Solution : (b) (b) 2 s ID (a) 1 s A stone dropped from a building of height h and it reaches after t seconds on earth. From the same YG building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t 1 and t 2 seconds respectively, then [CPMT 1997; UPSEAT 2002; KCET (Engg./Med.) 2002] (a) t  t1  t 2 For first case of dropping h  D Solution : (c) (b) t  t1  t 2 2 (c) t  U 1 g(t 2  t12 ) 2 ST 1 g(t 2  t 22 ) 2 1 2 1 2 gt 2  gt 2 2.......(ii) on solving these two equations :  Problem 62. 1 2 1 2 gt1  gt 2 2......(i) For third case of upward throwing h  ut 2   ut 2  (d) t  t12 t 22 1 2 gt. 2 For second case of downward throwing h  ut 1    ut 1  t1 t 2 t1 t 2  t12  2  t t2 t  t 22 t1 t 2. By which velocity a ball be projected vertically downward so that the distance covered by it in 5th second is twice the distance it covers in its 6th second ( g  10 m / s 2 ) (a) 58.8 m / s Solution : (c) (b) 49 m / s (c) 65 m / s 1 10 10 g (2n  1)  u  [2  5  1]  2 {u  [2  6  1]} 2 2 2  u  45  2  (u  55)  u  65 m / s. By formula hn  u  (d) 19.6 m / s 126 Motion In One Dimension Problem 63. Water drops fall at regular intervals from a tap which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant [CBSE PMT 1995] Solution : (b) (b) 3.75 m (c) 4.00 m Let the interval be t then from question 1 1.....(i) For second drop x  gt 2 g(2 t) 2  5 2 2 5 5 By solving (i) and (ii) x  and hence required height h  5   3.75 m. 4 4 Problem 64......(ii) E3 For first drop (d) 1.25 m 60 (a) 2.50 m A balloon is at a height of 81 m and is ascending upwards with a velocity of 12 m / s. A body of 2 kg weight is dropped from it. If g  10 m / s 2 , the body will reach the surface of the earth in (d) 6.75 s As the balloon is going up we will take initial velocity of falling body  12m / s, h  81m, g  10 m / s 2 By applying h  ut  1 2 1 gt ; 81  12 t  (10 )t 2  5 t 2  12t  81  0 2 2 12  144  1620 10  12  1764  5.4 sec. 10 YG  t Problem 65. (c) 5.4 s U Solution : (c) (b) 4.025 s ID (a) 1.5 s A particle is dropped under gravity from rest from a height h (g  9.8 m / s 2 ) and it travels a distance 9h/25 in the last second, the height h is (a) 100 m (b) 122.5 m [MNR 1987] (c) 145 m 1 2 gn = h 2.....(i) Distance travelled in n th sec  g 9h (2n  1)  2 25.....(ii) D Distance travelled in n sec = U Solution : (b) [MP PMT 1994] (d) 167.5 m Solving (i) and (ii) we get. h  122.5 m. A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. ST Problem 66. The height of the tower is (a) 3u 2 / g Solution : (b) (b) 4 u 2 / g (c) 6 u 2 / g For vertical downward motion we will consider initial velocity = – u. By applying v 2  u 2  2 gh , (3u) 2  (u) 2  2 gh  h  Problem 67. (d) 9 u 2 / g 4u 2. g A stone dropped from the top of the tower touches the ground in 4 sec. The height of the tower is about [MP PET 1986; AFMC 1994; CPMT 1997; BHU 1998; DPMT 1999; RPET 1999] Motion In One Dimension 127 (a) 80 m (b) 40 m (c) 20 m (d) 160 m 1 2 1 gt   10  4 2  80 m. 2 2 h Problem 68. A body is released from a great height and falls freely towards the earth. Another body is released from 60 Solution : (a) the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is (c) 19.6 m (d) 24.5 m The separation between two bodies, two second after the release of second body is given by : 1 1 g(t 12  t 22 ) =  9.8  (3 2  2 2 )  24.5 m. 2 2 s= ID Solution : (d) (b) 9.8 m E3 (a) 4.9 m 2.12 Motion With Variable Acceleration. (i) If acceleration is a function of time  t    f (t) dt dt U then v  u  a  f (t) 0 f (t) dt and s  ut  YG (ii) If acceleration is a function of distance then v 2  u 2  2 a  f (x )  x x0 f (x ) dx (iii) If acceleration is a function of velocity a = f ( v)  dv and x  x 0  f (v) v D then t  u  v u vdv f (v) U Sample problems based on variable acceleration Problem 69. An electron starting from rest has a velocity that increases linearly with the time that is v  kt, where ST k  2 m / sec 2. The distance travelled in the first 3 seconds will be (a) 9 m Solution : (a) x  t2 t1 Problem 70. (b) 16 m (d) 36 m 3 t 2  2t dt  2    9 m.  2  0 0 3 v dt  (c) 27 m [NCERT 1982]  The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v 0. The distance travelled by the particle in time t will be (a) v 0 t  Solution : (c)  v2 v1 1 2 bt 3 dv =  t2 t1 (b) v 0 t  a dt =  t2 t1 (bt ) dt 1 3 bt 3 (c) v 0 t  1 3 bt 6 [CBSE PMT 1995] (d) v 0 t  1 2 bt 2 128 Motion In One Dimension 2    t1  bt 2  v 2  v 1    2 2    v 0  bt  2 0 v 0 dt   t 60  S  bt 2 1 dt  v 0 t  bt 3 2 6 The motion of a particle is described by the equation u  at. The distance travelled by the particle in the E3 Problem 71. t  bt 2  v 2  v1    2  first 4 seconds Solution : (d) (b) 12a u  at  ds  at dt 4 t2   s  at dt  a   = 8a.  2  0 0 4 ST U D YG U  (c) 6a ID (a) 4a (d) 8 a [DCE 2000]