CBSE New Pattern Motion in a Straight Line Quick Revision PDF

CBSE New Pattern Motion in a Straight Line Quick Revision 1. Rest If the position of an object does not change Two-dimensional Motion (2-D) The w.r.t. its surrounding with the pa...

CBSE New Pattern Motion in a Straight Line Quick Revision 1. Rest If the position of an object does not change Two-dimensional Motion (2-D) The w.r.t. its surrounding with the passage of time, it is motion of an object is considered as 2-D, said to be at rest. e.g. Book lying on the table, a if two coordinates are needed to specify person sitting on a chair, etc. the position of the object. In 2-D motion, 2. Motion If the position of an object is continuously the object moves in a plane. e.g. A changing w.r.t. its surrounding w.r.t time, then it is said satellite revolving around the earth. to be in the state of motion. e.g. The crawling insects, Three-dimensional Motion (3-D) The water flowing down a dam, etc. motion of an object is considered as 3-D, 3. Types of Motion if all the three coordinates are needed to On the basis of the nature of path followed, motion specify the position of the object. is classified as This type of motion takes place in Rectilinear Motion The motion in which a three-dimensional space. particle moves along a straight line is called e.g. Butterfly flying in garden, the rectilinear motion. e.g. Motion of a sliding body motion of water molecules and motion on an inclined plane. of kite in the sky. Circular Motion The motion in which a particle 4. Point Object An object is considered as moves in a circular path is called circular motion. point object, if the size of the object is e.g. A string whirled in a circular loop. much smaller than the distance travelled Oscillatory Motion The motion in which a by it in a reasonable duration of time. particle moves to and fro about a given point is e.g. Earth can be considered as a point known as oscillatory motion. e.g. Simple object in its orbit. pendulum. 5. Position It is defined as the point where On the basis of the number of coordinates required an object is situated. to define the motion of an object, motion is 6. Path Length or Distance The length of classified as the path covered by the object in a given One-dimensional Motion (1-D) The motion of time-interval is known as its path length or an object is considered as 1-D, if only one distance travelled. It is a scalar quantity, coordinate is needed to specify the position of the i.e. it has only magnitude but no direction. object.  Ds 7. Displacement The change in position of an v i = Dlim t ®0 object in a particular direction is termed as Dt displacement, i.e. the difference between the ds final and initial positions of the object in a or vi = dt given time. It is denoted by Dx. Mathematically, it is represented by where, ds is the distance covered in time dt. Dx = x2 - x1 11. Velocity The rate of change in position or where, x 1 and x 2 are the initial and final displacement of an object with time is called positions of the object, respectively. the velocity of that object. Cases Displacement i.e. Velocity = Time 2 > x 1, then Dx is positive. If x It is a vector quantity. 1 > x 2, then Dx is negative. If x The velocity of an object can be positive, zero 1 = x 2, then Dx is zero. If x and negative according to its displacement. It is a vector quantity as it possesses both, the magnitude and direction. Unit of velocity In CGS, the unit of velocity 8. Uniform Motion in a Straight Line A body is cms -1 and in MKS or SI, it is ms -1. is said to be in a uniform motion, if it travels Dimensional formula [M 0LT -1 ] equal distances in equal intervals of time along Average Velocity Average velocity of a a straight line. A distance (x)-time (t) graph for body is defined as the change in position or uniform motion is a straight line passing displacement ( D x ) divided by the time through the origin. interval ( Dt ) in which that displacement 9. Non-uniform Motion A body is said to be in occurs. non-uniform motion, if it travels unequal Average velocity, displacements in equal intervals of time. Total displacement ( Dx ) 10. Speed The path length or the distance v av = Total time taken ( Dt ) covered by an object divided by the time taken to cover that distance is called its speed. Instantaneous Velocity Velocity at an Distance travelled instant is defined as the limit of average Speed = velocity as the time interval ( Dt ) becomes Time taken infinitesimally small or approaches to zero. It is a scalar quantity. The speed of the object Mathematically, instantaneous velocity (v i ) at for a given interval of time is always positive. an instant of time (t) is given by Unit of speed In SI (MKS) system, the unit of Dx speed is ms –1 and in CGS, it is cms -1. v i = lim Dt ® 0 D t Dimensional formula [M 0 LT -1] dx Average Speed Average speed of an or vi = dt object is defined as the total distance where, dx is displacement for time dt. travelled divided by the total time taken. Total distance travelled 12. Acceleration Acceleration of a body can be Average speed, v av = Total time taken expressed as the rate of change of velocity with Instantaneous Speed Speed at an instant is time. defined as the limit of the average speed as Change in velocity Acceleration = the time interval ( Dt ) becomes infinitesimally Time taken small or approaches to zero. It is a vector quantity. The SI unit of Mathematically, instantaneous speed (v i ) at acceleration is ms –2 and in CGS system, its unit any instant of time (t) is expressed as is cm s –2. Its dimensional formula is [M 0LT –2 ]. 13. Types of Acceleration 15. Non-uniformly Accelerated Motion Uniform Acceleration If an object is When acceleration of an object is not constant moving with uniform acceleration, it means or acceleration is a function of time, then that the change in velocity is equal in equal following relations hold for one-dimensional intervals of time. motion Non-uniform Acceleration If an object has v = dx variable or non-uniform acceleration, it dt means that, the change in velocity is unequal dx = v dt in equal intervals of time. a = dv dv Average Acceleration The average =v dt dx acceleration over a time interval is defined as dv = a dt or vdv = adx the change in velocity divided by the time interval. 16. Equations of Motion for the Motion of an Object under Gravity Average acceleration, Dv v 2 - v 1 When an object is thrown upwards or fall a av = = towards the earth under the effect of gravity Dt t2 - t1 only, then its motion is called motion under Instantaneous Acceleration It is defined gravity. as the acceleration of a body at a certain In this case, the equations of motion are given instant or the limiting value of average below acceleration when time interval becomes v = u + (+ g ) t very small or tends to zero. So, instantaneous Upward acceleration, motion 1 g h = ut + (+ g ) t2 Dv d v g 2 a inst = lim = Downward Dt ® 0 D t dt v 2 = u2 + 2(+ g )h motion dv In case of upward motion, acceleration due to where, is the differential coefficient of v gravity, g is taken as negative and for dt w.r.t. t. downward motion, g is taken as positive. 14. Kinematic Equations for 17. Stopping Distance for a Vehicle When Uniformly Accelerated Motion If the brakes are applied to a moving vehicle, the change in velocity of an object in each unit of distance it travels before stopping is called time is constant, then the object is said to be stopping distance. moving with constant acceleration and such a u2 motion is called uniformly accelerated Stopping distance, ds = 2a motion. An object moves along a straight line with a constant acceleration a and u be the where, u = initial velocity of the vehicle initial velocity at t = 0 and v be the final velocity and a = retardation. of the object after time (t), then 18. Relative Velocity in 1-D It is defined as the Velocity-Time Relation v = u + at time rate of change of relative position of one 1 2 object w.r.t. to another. Position-Time Relation x = ut + at 2 If an object A is moving with velocity v A and an where, x is the position of the object at time t. object B is moving with velocity v B , then the Position-Velocity Relation v 2 = u 2 + 2ax velocity of object A relative to object B is Displacement of the Object in given as v AB = v A - v B a The relative velocity of object B relative to nth Second s (nth ) = u + ( 2n - 1) object A is v BA = v B - v A 2 19. Different Graphs related to Motion are as Condition Graph follows Velocity Body moving with a Displacement-Time Graph constant retardation and A its initial velocity is Condition Graph non-zero Displacement For a stationary body B O Time Body moving with a O constant retardation with O zero initial velocity Time Body moving with a Displacement constant velocity Body moving with Velocity increasing acceleration O Time Displacement Body moving with a O constant acceleration Time Body moving with Velocity decreasing acceleration O Time Body moving with a Displacement constant retardation O Time O Time Note Slope of velocity-time graph gives average Body moving with Displacement acceleration. infinite velocity, but such B Acceleration-Time Graph motion of a body is never possible. Condition Graph Acceleration O A Time Body moving with a constant acceleration Note Slope of displacement-time graph gives average velocity. Velocity-Time Graph O Time Condition Graph Body moving with Acceleration Velocity Body moving with a constant increasing constant velocity acceleration O Time O Time Body moving with Acceleration Body moving with a Velocity constant decreasing constant acceleration acceleration having zero initial velocity O Time O Time CBSE New Pattern ~ Physics 11th (Term-I) 33 Objective Questions Multiple Choice Questions 5. The displacement of a car is given as 1. Which of the following is an example of - 240 m, here negative sign indicates one-dimensional motion? (a) direction of displacement (a) Landing of an aircraft (b) negative path length (b) Earth revolving around the sun (c) position of car at that point (c) Motion of wheels of moving train (d) no significance of negative sign (d) Train running on a straight track 6. Snehit starts from his home and walks 2. The coordinates of object with respect 50 m towards north, then he turns to a frame of reference at t = 0 s are towards east and walks 40 m and then ( - 1, 0 , 3). If t = 5 s, its coordinates are reaches his school after moving 20 m ( - 1, 0 , 4 ), then the object is in towards south. Then, his displacement (a) motion along Z-axis from his home to school is (b) motion along X-axis (a) 50 m (b) 110 m (c) motion along Y-axis (c) 80 m (d) 40 m (d) rest position between t = 0 s and t = 5 s 7. A vehicle travels half the distance l with 3. A person moves towards east for 3 m, speed v 1 and the other half with speed then towards north for 4 m and then v 2 , then its average speed is moves vertically up by 5 m. What is his (NCERT Exemplar) distance now from the starting point? v1 + v2 2v 1 + v 2 2v 1v 2 l (v 1 + v 2 ) (a) (b) (c) (d) (a) 5 2 m (b) 5 m (c) 10 m (d) 20 m 2 v1 + v2 v1 + v2 v 1v 2 4. For a stationary object at x = 40 m, the 8. A runner starts from O and comes back position-time graph is to O following path OQRO in 1h. What x (m) is his net displacement and average 40 speed? R (a) 20 0 t (s) 10 20 30 40 O Q x (m) 1km 40 (b) (a) 0,3.57 km/h (b) 0,0 km/h t (s) 0 20 (c) 0,2.57 km/h (d) 0,1 km/h x (m) 9. The sign ( + ve or - ve) of the average 40 velocity depends only upon (c) (a) the sign of displacement t (s) (b) the initial position of the object 30 (c) the final position of the object (d) None of the above (d) None of the above 10. Find the average velocity, when a (a) v1 >v2 particle completes the circle of radius (b) v2 >v1 (c) v1 =v2 1m in 10 s. (d) Data insufficient (a) 2 m/s (b) 3.14 m/s (c) 6.28 m/s (d) zero 14. For the x-t graph given below, the v - t 11. The displacement-time graph of two graph is shown correctly in moving particles make angles of 30° and 45° with the X -axis. The ratio of their velocities is x(m) Displacement (x) t(s) v (ms–1) (a) 45° 30° 0 0 t(s) Time (t) (a) 1 : 3 (b) 1 :2 (c) 1 :1 (d) 3 :2 v (ms–1) (b) 12. In figure, displacement-time (x - t ) graph 0 t(s) given below, the average velocity between time t = 5 s and t = 7 s is v (ms–1) (c) 35 30 x (m) 27.4 P2 0 t(s) 25 20 v (ms–1) 15 (d) 10 P1 5 0 t(s) 0 0 1 2 3 4 5 6 7 8 t (s) 15. The speed-time graph of a particle moving along a fixed direction is as (a) 8 ms-1 (b) 8.7 ms-1 shown in the figure. The distance (c) 7.8 ms-1 (d) 13.7 ms-1 traversed by the particle between t = 0 s to t = 10 s is 13. Figure shows the x-t plot of a particle in one-dimensional motion. Two different equal intervals of time show speed in time intervals 1 and 2 respectively, then v (ms–1) 12 A x C B O t (s) 2 5 10 t 1 (a) 20 m (b) 40 m (c) 60 m (d) 80 m 16. If an object is moving in a straight line, then x (m) (a) the directional aspect of vector can be specified by + ve and - ve signs (b) instantaneous speed at an instant is equal t1 to the magnitude of the instantaneous t (s) velocity at that instant (a) zero (c) Both (a) and (b) (b) positive (d) Neither (a) nor (b) (c) Data insufficient (d) Cannot be determined 17. In one dimensional motion, instantaneous speed v satisfies 22. A particle moves in a straight line. It 0 £ v < v 0. Then (NCERT Exemplar) can be accelerated (a) displacement in timeT must always take (a) only, if its speed changes by keeping its non-negative values direction same (b) displacement x in timeT satisfies (b) only, if its direction changes by keeping its - v 0 T < x < v 0T speed same (c) acceleration is always a non-negative (c) Either by changing its speed or direction number (d) None of the above (d) motion has no turning points 23. An object is moving along the path 18. The x-t equation is given as x = 2t + 1. OABO with constant speed, then The corresponding v-t graph is B (a) a straight line passing through origin (b) a straight line not passing through origin (c) a parabola O A (d) None of the above (a) the acceleration of the object while moving 19. The displacement x of an object is given along to path OABO is zero as a function of time, x = 2t + 3t 2. The (b) the acceleration of the object along the instantaneous velocity of the object at path OA and BO is zero t = 2 s is (c) there must be some acceleration along the (a) 16 ms -1 (b) 14 ms -1 path AB (c) 10 ms -1 (d) 12 ms -1 (d) Both (b) and (c) 20. The displacement of a particle starting 24. The average velocity of a body moving from rest (at t = 0) is given by with uniform acceleration travelling a s = 6t 2 - t 3. The time in seconds at distance of 3.06 m is 0.34 ms -1. If the which the particle will attain zero change in velocity of the body is velocity again is 0.18 ms -1 during this time, its uniform (a) 2 (b) 4 (c) 6 (d) 8 acceleration is (a) 0.01 ms-2 21. A car moves along a straight line (b) 0.02 ms-2 according to the x-t graph given below. (c) 0.03 ms-2 The instantaneous velocity of the car at (d) 0.04 ms-2 t = t 1 is 25. The slope of the straight line 30. The resulting a-t graph for the given v-t connecting the points corresponding to graph is correctly represented in (v 2 , t 2 ) and (v 1 , t 1 ) on a plot of velocity versus time gives 30 25 (a) average velocity 24 v(m/s) (b) average acceleration 20 15 (c) instantaneous velocity 10 (d) None of the above 5 0 26. The displacement x of a particle at time 2 4 6 8 10 12 1416 18 20 t(s) t along a straight line is given by x = a - bt + gt 2. The acceleration of the 4.8 2.4 particle is a (ms–2) 0 (a) – 2.4 2 4 6 8 10 12 14 16 18 20 22 (a) -b (b) - b + 2 g (c) 2 g (d) -2 g – 4.8 t (s) – 7.2 – 9.6 27. The displacement (in metre) of a –12.0 particle moving along X -axis is given by x = 18 t + 5t 2. The average 4.8 a (ms–2) 2.4 acceleration during the interval t 1 = 2 s (b) 0 2 4 6 8 10 12 14 16 18 20 22 and t 2 = 4 s is – 2.4 – 4.8 (a) 13 ms-2 (b) 10 ms-2 t (s) (c) 27 ms-2 (d) 37 ms-2 9.6 28. The relation between time and distance a (ms–2) is t = ax + bx , where a and b are 2 (c) 4.8 constants. The retardation is 2.4 (a) 2 av 3 (b) 2bv 3 0 2 4 6 8 10 12 (c) 2 abv 3 (d) 2b 2v 3 t (s) 29. The v -t graph of a moving object is shown in the figure. The maximum a(ms–2) acceleration is (d) 0 2 4 6 8 10 12 14 16 18 20 22 Velocity (cms−1) 80 t (s) 60 31. The kinematic equations of rectilinear 40 motion for constant acceleration for a 20 general situation, where the position coordinate at t = 0 is non-zero, say x 0 is 10 20 30 40 50 60 70 80 Time (s) (a) v = v 0 + at 1 2 (b) x = x0 + v 0 t + at (a) 1 cms -2 (b) 2 cms -2 2 -2 (c) v = v 0 + 2a (x - x0 ) 2 2 (c) 3 cms (d) 6 cms-2 (d) All of the above 32. The given acceleration-time graph 36. A particle is situated at x = 3 units at represents which of the following t = 0. It starts moving from rest with a physical situations? constant acceleration of 4 ms -2. The a position of the particle at t = 3 s is (a) x = + 21 units (b) x = + 18 units (c) x = -21 units (d) None of these 37. Consider the relation for relative t velocities between two objects A and B, (a) A cricket ball moving with a uniform speed v BA = - v AB is hit with a bat for a very short time interval. The above equation is valid, if (b) A ball is falling freely from the top of a (a) v A and v B are average velocities tower. (b) v A and v B are instantaneous velocities (c) A car moving with constant velocity on a (c) v A and v B are average speed straight road. (d) Both (a) and (b) (d) A football is kicked into the air vertically upwards. 38. A person is moving with a velocity of 33. An object is moving with velocity 10 m s -1 towards north. A car moving 10 ms -1. A constant force acts for 4 s on with a velocity of 20 ms -1 towards south the object and gives it a speed of 2 ms -1 crosses the person. in opposite direction. The acceleration The velocity of car relative to the produced is person is (a) 3 ms-2 (b) - 3 ms-2 (a) - 30 ms-1 (b) + 20ms-1 (c) 6 ms-2 (d) - 6 ms-2 (c) 10 ms-1 (d) - 10 ms-1 34. All the graphs below are intended to 39. A motion of a body is said to be ……, if represent the same motion. One of it moves along a straight line in any them does it incorrectly. Pick it up. direction. Velocity Distance (a) one-dimensional (b) two dimensional (a) Position (b) Time (c) three-dimensional (d) All of the above Position Velocity 40. The numerical ratio of displacement to the distance covered by an object is (c) Time (d) Time always equal to or less than ……. (a) 1 (b) zero (c) Both (a) and (b) (d) infinity 35. Velocity of a body moving along a 41. The time taken by a 150 m long train to straight line with uniform acceleration a cross a bridge of length 850 m is 80 s. It reduces by (3/4)th of its initial velocity is moving with a uniform velocity of in time t 0. The total time of motion of …… km/h. the body till its velocity becomes zero is (a) 45 (b) 90 4 3 5 8 (c) 60 (d) 70 (a) t0 (b) t0 (c) t0 (d) t0 3 2 3 3 42. The distance-time graph of …… is a With reference to the graph, which of straight line. the given statement(s) is/are incorrect? (a) uniform motion (a) The instantaneous speed during the (b) non-uniform motion interval t = 5 s to t = 10 s is negative at all (c) uniform acceleration time instants during the interval. (d) None of the above (b) The velocity and the average velocity for the interval t = 0 s to t = 5 s are equal and 43. Which of the following statement is positive. correct? (c) The car changes its direction of motion at t = 5 s. (a) The magnitude of average velocity is the (d) The instantaneous speed and the average speed. instantaneous velocity are positive at all (b) Average velocity is the displacement time instants during the interval t = 0 s to divided by time interval. t = 5 s. (c) When acceleration of particle is constant, then motion is called as non-uniformly 46. A graph of x versus t is shown in figure. accelerated motion. Choose correct statement given below. (d) When a particle returns to its starting point, x its displacement is non-zero. 44. For motion of the car between t = 18 s A B E and t = 20 s, which of the given C statement is correct? 296 D t x (m) 250 (a) The particle having some initial velocity at t = 0. 100 (b) At point B, the acceleration a > 0. (c) At point C, the velocity and the acceleration O 10 18 20 t(s) vanish. (d) The speed at E exceeds that at D. (a) The car is moving in a positive direction with a positive acceleration. 47. Match the Column I with Column II (b) The car is moving in a negative direction and select the correct option from the with a positive acceleration. codes given below (c) The car is moving in positive direction with a negative acceleration. Column I Column II (d) The car is moving in negative direction with a negative acceleration. A. d v / dt p. Acceleration 45. The x-t graph for motion of a car is B. d | v |/dt q. Rate of change of given below speed 10 dr x (m) C. r. Velocity dt d | r| Magnitude D. s. O 5 10 dt of velocity t(s)  Codes Column I Column II A B C D Position-time (a) p q r s graph of two A. p. (b) p r s q objects with x (m) (c) q p r s equal velocities. O t(s) (d) s r p q 48. Given x-t graph represents the motion Position-time of an object. Match the Column I B. graph of two q. (parts of graph) with Column II objects with x (m) (representation) and select the correct unequal option from the codes given below. velocities but O t(s) in same A direction. x Position-time C C. graph of two r. B objects with x (m) velocities in O opposite O t(s) t direction. Column I Column II Codes A. Part OA of p. Positive A B C A B C graph velocity (a) p q r (b) q p r B. Part AB of q. Object at rest (c) p r q (d) q r p graph C. Part BC of r. Negative Assertion-Reasoning MCQs graph velocity For question numbers 50 to 63, two D. Point A in the s. Change in statements are given-one labelled graph direction of Assertion (A) and the other labelled motion Reason (R). Select the correct answer to Codes these questions from the codes (a), (b), (c) A B C D and (d) are as given below (a) Both A and R are true and R is the (a) p q r s correct explanation of A. (b) p r q s (b) Both A and R are true but R is not the (c) q p r s correct explanation of A. (d) s r q p (c) A is true but R is false. (d) A is false and R is also false. 49. Match the Column I (position-time graph) with Column II (representation) 50. Assertion In real-life, in a number of and select the correct option from the situations, the object is treated as a codes given below. point object.  Reason An object is treated as point Reason Infinite acceleration cannot be object, as far as its size is much smaller realised in practice. than the distance, it moves in a reasonable duration of time. 58. Assertion In realistic situation, the x-t, v-t and a-t graphs will be smooth. 51. Assertion If the displacement of the body is zero, the distance covered by it Reason Physically acceleration and may not be zero. velocity cannot change values abruptly at an instant. Reason Displacement is a vector quantity and distance is a scalar quantity. 59. Assertion A body cannot be accelerated, when it is moving 52. Assertion An object can have constant uniformly. speed but variable velocity. Reason When direction of motion of Reason SI unit of speed is m/s. the body changes, then body does not have acceleration. 53. Assertion The speed of a body can be negative. 60. Assertion For uniform motion, velocity Reason If the body is moving in the is the same as the average velocity at all opposite direction of positive motion, instants. then its speed is negative. Reason In uniform motion along a straight line, the object covers equal 54. Assertion For motion along a straight distances in equal intervals of time. line and in the same direction, the magnitude of average velocity is equal 61. Assertion A body is momentarily at to the average speed. rest at the instant, if it reverse the Reason For motion along a straight line direction. and in the same direction, the Reason A body cannot have magnitude of displacement is not equal acceleration, if its velocity is zero at a to the path length. given instant of time. 55. Assertion An object may have varying 62. Assertion In the s-t diagram as shown speed without having varying velocity. in figure, the body starts moving in positive direction but not from s = 0. Reason If the velocity is zero at an instant, the acceleration is zero at that s instant. 56. Assertion Acceleration of a moving t particle can change its direction without t0 any change in direction of velocity. Reason If the direction of change in velocity vector changes, direction of acceleration vector does not changes. Reason At t = t 0 , velocity of body 57. Assertion The v-t graph perpendicular changes its direction of motion. to time axis is not possible in practice.  63. Assertion If acceleration of a particle 67. If the car goes from O to P and returns moving in a straight line varies back to O, the displacement of the as a µ t n , then s µ t n + 2. journey is Reason If a-t graph is a straight (a) zero (b) 720 m (c) 420 m (d) 340 m line,then s-t graph may be a parabola. 68. The path length of journey from O to P Case Based MCQs and back to O is Direction Answer the questions from (a) 0 m (b) 720 m 64-68 on the following case. (c) 360 m (d) 480 m Motion in a Straight Line Direction Answer the questions from 69-73 on the following case. If the position of an object is continuously changing w.r.t. its surrounding, then it is said Average Speed and Average Velocity to be in the state of motion. Thus, motion can When an object is in motion, its position be defined as a change in position of an object changes with time. So, the quantity that with time. It is common to everything in the describes how fast is the position changing universe. w.r.t. time and in what direction is given by In the given figure, let P, Q and R represent average velocity. the position of a car at different instants of It is defined as the change in position or time. displacement (Dx ) divided by the time interval R O Q P (Dt ) in which that displacement occur. 80 120 160 200 240 280 320 360 400 (m) –160 –120 –80 –40 0 40 X-axis However, the quantity used to describe the rate of motion over the actual path, is average 64. With reference to the given figure, the speed. It defined as the total distance travelled position coordinates of points P and R by the object divided by the total time taken. are (a) P º (+ 360, 0, 0); R º (- 120, 0, 0) 69. A 250 m long train is moving with a (b) P º (- 360, 0, 0); R º (+ 120, 0 , 0) uniform velocity of 45 kmh - 1. The time taken by the train to cross a bridge of (c) P º (0, + 360, 0); R º (- 120, 0, 0) length 750 m is (d) P º (0, 0, + 360); R º (0, 0, - 120) (a) 56 s (b) 68 s 65. Displacement of an object can be (c) 80 s (d) 92 s (a) positive (b) negative 70. A truck requires 3 hr to complete a journey of 150 km. What is average (c) zero (d) All of the above speed? (a) 50 km/h (b) 25 km/h 66. The displacement of a car in moving (c) 15 km/h (d) 10 km/h from O to P and its displacement in moving from P to Q are 71. Average speed of a car between points A and B is 20 m/s, between B and C is (a) + 360 m and - 120 m (b) - 120 m and + 360 m 15 m/s and between C and D is 10 m/s. (c) + 360 m and + 120 m What is the average speed between A (d) + 360 m and - 600 m and D, if the time taken in the  mentioned sections is 20s, 10s and 5s, The motion in which the acceleration remains respectively? constant is known as to be uniformly (a) 17.14 m/s (b) 15 m/s accelerated motion. There are certain (c) 10 m/s (d) 45 m/s equations which are used to relate the displacement (x), time taken (t ), initial velocity 72. A cyclist is moving on a circular track of radius 40 m completes half a (u ), final velocity (v ) and acceleration (a ) for revolution in 40 s. Its average velocity such a motion and are known as kinematics is equations for uniformly accelerated motion. (a) zero (b) 2 ms -1 74. The displacement of a body in 8 s (c) 4 p ms-1 (d) 8 p ms-1 starting from rest with an acceleration 73. In the following graph, average velocity of 20 cms -2 is is geometrically represented by (a) 64 m (b) 640 m (c) 64 cm (d) 0.064 m 35 30 75. A particle starts with a velocity of x (m) 27.4 P2 2 ms -1 and moves in a straight line with 25. ms -2. The first time a retardation of 01 20 15 at which the particle is 15 m from the 10 P1 starting point is 5 (a) 10 s (b) 20 s 0 (c) 30 s (d) 40 s 0 1 2 3 4 5 6 7 8 t (s) 76. If a body starts from rest and travels (a) length of the line P1 P2 120 cm in 6th second, then what is its (b) slope of the straight line P1 P2 (c) slope of the tangent to the curve at P1 acceleration? (d) slope of the tangent to the curve at P2 (a) 0.20 ms- 2 (b) 0027. ms- 2. ms- 2 (c) 0218. ms- 2 (d) 003 Direction Answer the questions from 74-78 on the following case. 77. An object starts from rest and moves with uniform acceleration a. The final Uniformly Accelerated Motion velocity of the particle in terms of the The velocity of an object, in general, changes distance x covered by it is given as during its course of motion. Initially, at the (a) 2ax (b) 2ax time of Galileo, it was thought that, this ax (c) (d) ax change could be described by the rate of 2 change of velocity with distance. But, through his studies of motion of freely falling objects 78. A body travelling with uniform and motion of objects on an inclined plane, acceleration crosses two points A and B Galileo concluded that, the rate of change of with velocities 20 ms -1 and 30 ms -1 , velocity with time is a constant of motion for respectively. The speed of the body at mid-point of A and B is all objects in free fall. (a) 25 ms-1 (b) 25.5 ms-1 This led to the concept of acceleration as the (c) 24 ms-1 (d) 10 6 ms-1 rate of change of velocity with time. CBSE New Pattern ~ Physics 11th (Term-I) 43 ANSWERS Multiple Choice Questions 1. (d) 2. (a) 3. (a) 4. (a) 5. (a) 6. (a) 7. (c) 8. (a) 9. (a) 10. (d) 11. (a) 12. (b) 13. (b) 14. (a) 15. (c) 16. (c) 17. (b) 18. (b) 19. (b) 20. (b) 21. (a) 22. (c) 23. (d) 24. (b) 25. (b) 26. (c) 27. (b) 28. (a) 29. (d) 30. (a) 31. (d) 32. (a) 33. (b) 34. (b) 35. (a) 36. (a) 37. (d) 38. (a) 39. (a) 40. (a) 41. (a) 42. (a) 43. (b) 44. (a) 45. (a) 46. (c) 47. (a) 48. (b) 49. (b) Assertion-Reasoning MCQs 50. (a) 51. (b) 52. (b) 53. (d) 54. (c) 55. (d) 56. (d) 57. (a) 58. (a) 59. (d) 60. (b) 61. (c) 62. (c) 63. (b) Case Based MCQs 64. (a) 65. (d) 66. (a) 67. (a) 68. (b) 69. (c) 70. (a) 71. (a) 72. (b) 73. (b) 74. (c) 75. (a) 76. (c) 77. (a) 78. (b) SOLUTIONS 1. In one-dimensional motion, only one B (40 m) towards east and from B to C (20 m) towards south as shown in the figure below. coordinate is required to specify the position of the object. So, a train running on a straight 40 m N A B track is an example of one-dimensional motion. 40 m 20 m W E D 2. Given, at t = 0 s, position of an object is 50 m C S ( -1 , 0 , 3) and at t = 5 s, its coordinate is ( -1 , 0 , 4 ). So, there is no change in x and 30 m y-coordinates, while z -coordinate changes θ P from 3 to 4. So,the object is in motion along O Z-axis. Displacement of Snehit is OC, which can be 3. Distance from starting point calculated by Pythagoras theorem, i.e. = ( 3) 2 + ( 4 ) 2 + ( 5) 2 = 5 2 m In DODC, OC 2 = OD 2 + CD 2 = ( 30 ) 2 + ( 40 ) 2 4. For a stationary object, the position-time = 900 + 1600 = 2500 graph is a straight line parallel to the time Þ OC = 50 m axis, so for the given object at x = 40 m, x-t graph is correctly shown in option (a). 7. Time taken to travel first half distance, l/ 2 l 5. In I-D motion, positive and negative signs are t1 = = used to specify the direction of motion. v1 2v 1 Since, displacement is a vector quantity, so Time taken to travel second half distance, negative sign in -240 m indicates the l t2 = direction of displacement. 2v 2 6. Let O be the starting point, i.e. home. So, l l lé1 1ù Total time = t 1 + t 2 = + = ê + ú according to the question, Snehit moves from 2v 1 2v 2 2 ë v 1 v 2 û O to A (50 m) towards north, then from A to  We know that, v av = average speed 12. Given, x 2 = 27.4 m, x 1 = 10 m, t 2 = 7 s and total distance = t 1 = 5 s. total time Average velocity between 5 s and 7s, l 2v 1v 2 = = x - x 1 27.4 -10 lé1 1 ù v1 + v2 v = 2 = ê + ú t2 - t1 7-5 2 ëv1 v2 û 17.4 = = 8.7 ms -1 8. As runner starts from O and comes back 2 to O, so net displacement is zero. 13. Slope of x-t graph in a small interval Average speed = Average speed in that interval Total distance OQ + QR + RO As, slope for interval 2 > slope for interval 1. = = Total time Total time \ v2 > v1 æ 90° ö 1 km + ( 2pr ) ç ÷ km + 1 km 14. The x - t graph shown, is parallel to time axis. è 360° ø = This means that, the object is at rest. So, the 1h velocity of the object is zero for all time (Q angle of sector OQR is 90°) instants. Hence, v -t graph coincides with the æ1ö time axis as shown in graph (a). 1 + 2p ´ 1 ç ÷ + 1 è 4ø 15. Distance travelled by the particle between time = 1 interval t = 0 s to t = 10 s p = Area of triangle OAB = 2 + = 3.57 km/h 2 1 = ´ Base ´ Height 9. Since, average velocity, 2 D x Displacement 1 v = = = ´ OB ´ AC D t Time interval 2 1 So, average velocity depends on the = ´ 10 ´ 12 = 60 m 2 displacement and hence it depends on the sign of the displacement. 16. In one-dimensional motion, i.e. motion along a straight line, there are only two directions 10. When a particle completes one revolution in in which an object can move and these two circular motion, then average displacement directions can be easily specified by + ve and travelled by particle is zero. - ve signs. Hence, average velocity Also, in this motion instantaneous speed or average displacement 0 = = =0 simply speed at an instant is equal to the Dt Dt magnitude of instantaneous velocity at the 11. In case x - t graph is a straight line, the slope given instant. of this line gives velocity of the particle. 17. For maximum and minimum displacements, As slope = tan q, where q is the angle which we have to keep in mind the magnitude and the tangent to the curve makes with the direction of maximum velocity. horizontal in anti-clockwise direction. As, maximum velocity in positive direction is The velocities of two particles A and B are v 0 and maximum velocity in opposite 1 direction is also - v 0. v A = tan 30° = 3 Maximum displacement in one direction = v 0T v B = tan 45° = 1 Maximum displacement in opposite The ratio of velocities, directions = - v 0T 1 Hence, the range of displacement will be vA :vB = :1 = 1 : 3 3 -v 0T < x < v 0T.  dx Since, the direction of velocity is changing, 18. v = = 2 ms -1 = constant dt i.e. there must be some acceleration along the v ms–1 path AB. Distance 3.06 24. Time = = =9 s 2 Average velocity 0.34 Acceleration t (s) Change in velocity = Time Hence, option (b) is correct. 0.18 = = 0.02 ms -2 19. Given, x = 2t + 3t 2 9 dx 25. Average acceleration is defined as the v = = 2+ 6t dt average change of velocity per unit time. On For t = 2 s, v = 2 + 6 ( 2) = 14 ms -1 a plot of v -t, the average acceleration is the 20. Displacement of the particle, slope of the straight line connecting the points corresponding to ( v 2, t 2 ) and ( v 1, t 1 ). s = 6t 2 - t 3 Velocity of the particle, 26. Given, x = a - bt + g t 2 ds d dx d v = = ( 6t 2 - t 3 ) v = = ( a - bt + g t 2 ) = - b + 2 g t dt dt dt dt v = 12t - 3t 2 dv d a = = ( - b + 2g t ) = 2g For v = 0 Þ 12t = 3 t 2 Þ t = 4 s dt dt 21. The instantaneous velocity is the slope of the 27. Given, x = 18t + 5t 2 tangent to the x -t graph at that instant of dx d v = = (18t + 5t 2 ) = 18 + 10t time. dt dt Tangent at point P \ v = 10t + 18 P corresponding to At t 1 = 2 s, v 1 = 10 ( 2) + 18 = 38 m/s x (m) t = t1 At t 2 = 4 s, v 2 = 10 ( 4 ) + 18 = 58 m/s v -v 58 - 38 20 \ a = 2 1 = = = 10 ms -2 t = t1 t(s) t 2 2 28. Given, t = ax 2 + bx At t = t 1, the tangent is parallel to time axis as dt shown above and hence its slope is zero. = 2a x + b dx Thus, instantaneous velocity at t = t 1 is zero. dx 1 22. Since velocity is a vector quantity, having Þ =v = dt 2ax + b both magnitude and direction. So, a change in velocity may involve change in either or dv dv dx As, acceleration, a == both of these factors. Therefore, acceleration dt dx dt may result from a change in speed dv 1 æ - v 2a ö (magnitude), a change in direction or changes Þa =v × = ç ÷ dx 2ax + b è 2ax + b ø in both. = - 2a v × v 2 = - 2av 3 23. For paths OA and BO, the magnitude of \ Retardation = 2av 3 velocity (speed) and direction is constant, hence acceleration is zero. For path AB, since 29. Maximum acceleration means maximum this path is a curve, so the direction of the change in velocity in minimum time interval. velocity changes at every moment but the In time interval t = 30 s to t = 40 s, magnitude of velocity (speed) remains Dv 80 - 20 60 a = = = = 6 cms -2 constant. Dt 40 - 30 10 30. Average acceleration for different time 37. Given, v BA = - v AB intervals is the slope of v-t graph, which are as follows The above relation is true for both average ( 24 - 0 ) ms -1 velocities of particles and instantaneous For 0 s -10 s, a = = 2.4 ms -2 (10 - 0 ) s velocities of particles. ( 24 - 24 ) ms -1 As speed is scalar quantity, ignorant of For 10 s -18 s, a = = 0 ms -2 (18 - 10 ) s direction, so average speed may not be equal. ( 0 - 24 ) ms -1 38. Let south to north direction be positive. For 18 s - 20 s, a = = - 12 ms -2 ( 20 - 18 ) s Velocity of car, vC = - 20 ms -1 So, the corresponding a -t graph for the given Velocity of person, v P = + 10 ms -1 v -t graph is shown correctly in graph (a). vCP = vC - v P = ( - 20 ) - (10 ) = - 30 ms -1 31. All the equations given in options (a), (b) and 39. One-dimensional motion is a motion along (c) are the kinematic equations of rectilinear a straight line in any direction. e.g. A train is motion for constant acceleration. moving on a platform. Hence, option (a) is correct. 32. The acceleration-time graph represents the motion of a uniformly moving cricket ball 40. Since, displacement d is always less than or turned back by hitting it with a bat for a very equal to the distance D but never greater short time interval. than it, i.e. d £ D. So, numerical ratio of 33. Given, v = - 2 ms -1 (opposite direction), displacement to the distance covered by an t = 4 s and u = 10 ms -1 object is always equal to or less than one. \ v = u + at or - 2 = 10 + 4a or a = - 3 ms -2 41. Total distance = Length of train + Length of bridge 34. If velocity versus time graph is a straight line = (150 + 850 ) m = 1000 m with negative slope, then acceleration is constant and negative. Distance 1000 Time = Þ 80 = Velocity v With a negative slope, distance-time graph æ 1 ö 1000 1000 18 will be parabolic çs = ut - at 2 ÷. v = m/s Þ v = ´ = 45 km/h è 2 ø 80 80 5 Hence, options (a), (c) and (d) are correct, so 42. In uniform motion, the velocity of an object option (b) will be incorrect. does not change or it remains constant with time. 35. According to kinematic equation of motion, So, the graph of distance-time is a straight v = u - at line. 3u u where, v = u - = u /4 Þ = u - at 0 43. Statement given in option (b) is correct but 4 4 the rest are incorrect and these can be Negatve sign signifies that the body will corrected as, decelerate, since the final velocity is In general, average speed is not equal to decreasing. magnitude of average velocity. It can be so, if u 4 the motion is along a straight line without or = t0 a 3 change in direction. u 4 Now, 0 = u - at or t = = t0 When acceleration of particle is not constant, a 3 then motion is called as non-uniformly 36. Given, x 0 = 3 units, a = 4 ms -2, t = 3 s accelerated motion. 1 2 Displacement is zero, when a particle returns Using relation, x = x 0 + v 0 t + at 2 to its starting point. 1 =3+ ´ 4 ´ ( 3) 2 = + 21 units 2 44. For negative acceleration, the x - t graph dr is the magnitude of rate of change of moves downward. But the car is moving in dt positive direction as the position coordinate is position of particle. This means it represents increasing in the positive direction. magnitude of velocity. Thus, the statement given in option (a) is Hence, A ® p, B ® q, C ® r and D ® s. correct, rest are incorrect. 48. In x -t graph, OA ® Positive slope ® Positive 45. The instantaneous speed is always positive as velocity it is the magnitude of the velocity at an AB ® Negative slope ® Negative velocity instant, so it is positive during t = 5 s BC ® Zero slope ® Object is at rest to t = 10 s. At point A, there is a change in sign of For t = 0 s to t = 5 s, the motion is uniform velocity, hence the direction of motion must and x-t graph has positive slope. So, the have changed at A. velocity and average velocity, instantaneous Hence, A ® p, B ® r, C ® q and D ® s. velocity and instantaneous speed are equal and positive. 49. A. For equal velocities, the slope of the straight lines must be same as shown below During t = 0 s to t = 5 s, the slope of the graph is positive, hence the average velocity and the velocity both are positive. x θ During t = 5 s to t = 10 s, the slope of the θ (Equal) same slope graph is negative, hence the velocity is O t(s) negative. Since, there is a change in sign of velocity at t = 5 s, so the car changes its B. For unequal velocity, slope is different, but direction at this instant. since, the objects are moving in the same direction, the slope for both the graphs Hence, option (a) is incorrect, while all others must be of same sign (positive or negative) are correct. and they meet at a point as shown below 46. As, point A is the starting point, therefore particle is starting from rest. x At point B, the graph is parallel to time axis, θ2 so the velocity is constant here. Thus, θ1 acceleration is zero. Also point C, the graph changes slope, hence C. For velocities in opposite direction, slopes velocity also changes. must be of opposite sign. Slope = tan q, After graph at C is almost parallel to time where q is the angle of the straight line with axis, hence we can say that velocity and horizontal in anti-clockwise direction. As, acceleration vanishes. we know, tan q1 > 0, tan q2 < 0. From the graph, it is clear that Hence, slopes are of opposite sign. slope at D > slope at E This condition is shown below Hence, speed at D will be more than at E. dv 47. is the rate of change of velocity, so it x dt θ1 θ2 represents acceleration. d | v| O t(s) is rate of change of speed of the particle. dt Hence A ® q, B ® p and C ® r. dr is the rate by which distance of particle 50. The approximation of an object as point dt object is valid only, when the size of the from the origin is changing.  object is much smaller than the distance it dv 57. Acceleration, a = = Slope of v-t graph moves in a reasonable duration of time. dt Therefore, both A and R are true and R is It v-t graph is perpendicular to t-axis, slope the correct explanation of A. =¥ 51. Distance is the total path length travelled by \ a =¥ the object. But displacement the shortest distance between the initial and final Therefore, both A and R are true and R is positions of the object. So, distance can never the correct explanation of A. be negative or zero. But displacement can be 58. In realistic situation, the x - t , v - t and a - t zero, positive and negative. graphs will be smooth, as the values of Also, distance is a scalar quantity. It means acceleration and velocity cannot change that, it is always positive but however abruptly since changes are always continuous. displacement is a vector quantity. So, it may Therefore, both A and R are true and R is be positive, zero or negative depending on the correct explanation of A. given situation. 59. The uniform motion of a body means that, Therefore, both A and R are true but R is the body is moving with constant velocity. not the correct explanation of A. But if the direction of motion is changing 52. Velocity is a vector quantity, so it has both (such as in uniform circular motion), its direction and magnitude. Hence, an object velocity changes and thus uniform can have variable velocity by keeping its acceleration is produced in the body. magnitude constant, i.e. speed and by Therefore, A is false and R is also false. changing direction only. 60. In uniform motion along a straight line, the The SI unit of speed is m/s. object covers equal distances in equal Therefore, both A and R are true but R is intervals of time. not the correct explanation of A. For uniform motion, x - t graph is represented 53. Speed can never be negative because it is a as a straight line inclined to time axis. The scalar quantity. So, if a body is moving in average velocity during any time interval negative direction, then also the speed will be t = t 1 to t = t 2 is the slope of the line PQ positive. which coincides with the graph. Therefore, A is false and R is also false. Q 54. For motion in a straight line and in the same direction, P x Displacement = Total path length Þ Average velocity = Average speed Therefore, A is true but R is false. t1 t2 t 55. If speed varies, then velocity will definitely Also, velocity at any instant say t = t 1 is the vary. slope of the tangent at point P which again When a particle is thrown upwards, at coincides with PQ or with the graph. Hence, highest point a ¹ 0 but v = 0. velocity is same as the average velocity at all Therefore, A is false and R is also false. instants. v - v i dv Therefore, both A and R are true but R is 56. Accleration, a = f = , i.e. direction of Dt dt not the correct explanation of A. acceleration is same as that of change in 61. When a particle is released from rest position velocity vector or in the direction of Dv. under gravity, then v = 0 but a ¹ 0. Therefore, A is false and R is also false.  Also, a body is momentarily at rest at the 71. Total distance ( d = vt ) instant, if it reverse the direction. = 20 ´ 20 + 15 ´ 10 + 10 ´ 5 = 600 m Therefore, A is true but R is false. Total time = 20 + 10 + 5 = 35 s 62. Slope of s-t graph = velocity = positive Therefore, average speed At t = 0, s ¹ 0, further at t = t 0 : s = 0, v ¹ 0. = 600 / 35 = 1714. m/s Therefore, A is true but R is false. 72. Given, R = 40 m and t = 40 s 63. By differentiating a-t equation two times, we Displacement will get s-t equation. Average velocity = Time taken Further s 2R 2 ´ 40 a = = = 2 ms -1 t 40 73. From the position-time graph, average t t velocity is geometrically represented by the Straight line Parabola slope of curve, i.e. slope of straight line P1 P2. Therefore, both A and R are true but R is 1 74. Displacement, s = ´ (0.2) (64) = 64 cm not the correct explanation of A. 2 1 64. The position coordinates of point 75. From equation of motion, s = ut - at 2 P = ( +360, 0, 0 ) and point R = ( -120, 0, 0 ). 2 1 65. Displacement is a vector quantity, it can be 15 = 2t - ´ ( 0.1)t Þ t = 10 s 2 2 positive, negative and zero. 76. From equation of motion, 66. Displacement, Dx = x 2 - x 1 a sn = u + ( 2n - 1) For journey of car in moving from O to P , 2 x 2 = + 360 m a Þ 1.2 = 0 + ( 2 ´ 6 - 1) x1 = 0 2 Þ Dx = x 2 - x 1 = 360 - 0 = + 360 m 1.2 ´ 2 Þ a = = 0.218 ms - 2 For journey, of car in moving from P to Q , 11 x 2 = + 240 m 77. Given, v0 = 0 x 1 = + 360 m Using relation, v 2 = v 02 + 2ax Þ Dx = x 2 - x 1 = 240 - 360 = - 120 m v 2 = 2ax Here, -ve sign implies that the displacement is in –ve direction, i.e. towards left. \ v = 2ax 67. Displacement, D x = x 2 - x 1 = 0 - 0 = 0 78. Let the acceleration of the car = a and distance between A and B = d 68. Path length of the journey Given, v = 30 ms -1 and u = 20 ms -1 = OP + PO = + 360 m + ( + 360 ) m = 720 m 2ad = ( 30 ) 2 - ( 20 ) 2 Total distance 900 - 400 69. Total time taken = ad = = 250 Speed 2 250 + 750 When the car is at the mid-point of AB, then t= = 80 s 5 speed of car is v 1. 45 ´ 18 v 12 - ( 20 ) 2 = 2a ( d / 2) Total distance v 12 = ad + 400 70. Average speed = Total time = 250 + 400 = 650 150 = = 50 km/h Therefore, v 1 = 25.5 ms -1 3