Kinematics (Physics) PDF

## CHAPTER 2 KINEMATICS (RECTILINEAR, PROJECTILE & RELATIVE MOTION) For NSEJS ### MECHANICS Mechanics is the branch of physics which deals with the cause and effects of motion of a particle, rigid objects and deformable bodies etc. Mechanics is classified under two streams namely Statics and Dynamics. Dynamics is further divided into Kinematics and Kinetics. | Mechanics | | |----------------------|--------------------------------------------------------------------------------------------------------------| | Statics | Study of forces and their effect on objects at rest. | | Dynamics | Study of forces and their effect on objects in motion. | | Kinematics | The word kinematics means 'science of motion'. Branch of mechanics that deals with study of motion without going into the cause of motion, i.e force, torque etc. | | Kinetics | It is branch of mechanics that is concerned about the causes (i.e. the force, torque) that cause motion of bodies. | ### RECTILINEAR MOTION #### Rest and Motion ##### Definition of Rest and Motion Motion is a combined property of the object and the observer. There is no meaning of rest or motion without the observer. Nothing is in absolute rest or in absolute motion. An object is said to be in motion with respect to a observer, if its position changes with respect to that observer. It may happen by both ways either observer moves or object moves. - **Rest:** An object is said to be at rest if it does not change its position w.r.t. its surroundings with the passage of time. > Eg. The chair, black board, table in the class room are at rest w.r.t. the students. - **Motion:** A body is said to be in motion if its position changes continuously w.r.t. the surroundings (or with respect to an observer) with the passage of time. > Eg. A car moving on the road will be in motion w.r.t. to the person standing on the road Rest and motion are relative terms, there is nothing like absolute motion or rest. >> Eg. A train is moving on the track, the passengers are seated, will be stationary with respect to each other but in moving condition with respect to station. Therefore, all the motions are relative. There is nothing like absolute motion. ##### Uniform and Non-Uniform motion: - **Uniform Motion:** A body has uniform motion if it travels equal distances in equal intervals of time, no matter how small these time intervals may be. > For example, a car running at a constant speed say, $10$ metre per second, will cover equal distances of $10$ metre every second, so its motion will be uniform. Please note that the distance-time graph for uniform motion is a straight line. - **Non-Uniform Motion:** A body has a non-uniform motion if it travels unequal distances in equal intervals of time. > For example, if we drop a ball from the roof of a building, we will find that it covers unequal distances in equal intervals of time. >> It covers: $4.9$ metre in the $1$st second, $14.7$ metre in the $2$nd second, $24.5$ metre in the $3$rd second, and so on. Thus, a freely falling ball covers smaller distance in the initial '1 second' interval and larger distance in the later '1 second' interval. From this discussion we conclude that the motion of a freely falling body is an example of non-uniform motion. The motion of a train starting from the railway station is also an example of non-uniform motion. This is because when the train starts from a station, it moves a very small distance in the 'first' second. The train moves a little more distance in the '2nd' second and so on. And when the train approaches the next station, the distance travelled by it per second decreases. ##### Features of uniform motion: 1. The velocity in uniform motion does not depend on the choice of origin. 2. The velocity in uniform motion does not depend on the choice of the time interval. 3. For uniform motion along a straight line in the same direction, the magnitude of the displacement is equal to the actual distance covered by the object. 4. The velocity is positive if the object is moving towards the right of the origin and negative if the object is moving towards the left of the origin. 5. For an object in uniform motion no force is required to maintain its motion. 6. In uniform motion, the instantaneous velocity is equal to the average velocity at all times because velocity remains constant at each instant, at each point of the path. #### Distance and Displacement: 1. **Distance:** It is the actual length of path covered by a moving particle. It is a scalar quantity. Its S.I. unit is metre (m). 2. **Displacement:** It is the shortest distance between the initial and final position of the particle. It is a vector quantity. Its S.I. unit is metre (m). > Eg.: Consider a body moving from a point A to a point B along the path shown in figure. Then total length path covered is called distance (path-1). While the length of straight line AB in the direction from A to B is called displacement (path-2). >> If a body travels in such a way that it comes back to its starting position, then the displacement is zero. However, distance travelled is never zero in case of moving body. ##### Some important points: 1. When an object moves towards right from origin, its displacement consider as positive. 2. When an object moves towards left from origin its displacement consider as negative. 3. When an object remains stationary or it moves first towards right and then an equal distance towards left, its displacement is zero. 4. Shifting origin causes no change in displacement. 5. If body moves along the circumference of the circle of radius r then distance travelled by it is given by $2πr$ and displacement is given by zero, for one complete revolution. | Distance | Displacement | |-------------|--------------| | 1. Distance is the length of the path actually traveled by a body in any direction. | 1. Displacement is the shortest distance between the initial and the final positions of a body in the direction of the point of the final position. | | 2. Distance between two given points depends upon the path chosen. | 2. Displacement between two points is measured by the straight path between the points. | | 3. Distance is always positive. | 3. Displacement may be positive as well as negative and even zero. | | 4. Distance is a scalar quantity. | 4. Displacement is a vector quantity. | | 5. Distance will never decrease. | 5. Displacement may decrease. | #### Speed The distance travelled by a body in unit time is called its speed. > $$speed = \frac{Distance}{Time} \text{ or } s = \frac{d}{t}$$ S.I. unit of speed or average speed is m/sec. It is a scalar quantity. - **Uniform Speed (or Constant Speed):** When an object covers equal distance in equal intervals of time, it is said to move with uniform speed. > Eg. A car moves $10$ m in every one second so its motion is uniform. - **Variable Speed (Non-Uniform Speed):** If a body covers unequal distance in equal intervals of time, its motion is said to be non-uniform. > Eg. Falling of an apple from a tree, a cyclist moving on a rough road, an athlete running a race, vehicle starting from rest, the motion of a freely falling body etc. - **Average Speed:** For an object moving with variable speed, it is the total distance travelled by the object divided by the total time taken to cover that distance. >1. $$Average\ speed = \frac{total\ distance\ travelled}{total\ time\ taken}$$ >> Let initial speed of an object is $v1$, final speed is $v2$ and acceleration is constant, then >>> $$average\ speed = \frac{V1 + V2}{2}$$ >2. A body covers a distance $s₁$ in time $t1$, $s2$ in time $t2$ and $s3$ in time $t3$. >> $$Then,\ average\ speed,\ V_{av}=\frac{S1+S2 + S3}{t₁+t2+t3}$$ >3. A body travels with speed $v₁$ for a time $t₁$ $v2$ for time $t2$ and $v3$ for the time $t3$. >> $$Then,\ average\ speed,\ V_{av}=\frac{V1t1+V2t2 + V3t3}{t₁+t2+t3}$$ >> $$.. S1 = V1t1, S2 = V2t2\ and\ S3 = V3t3$$ >> $$if\ t₁=t2=t3 = t$$ >>> $$V_{av} = \frac{t(V1 + V2 + V3)}{3t}$$ >>> $$V_av= \frac{(V1 + V2 + V3)}{3}$$ >4. A body covers a distance $s₁$ with speed $v1$, $S2$ with speed $v2$ and $s3$ with speed $v3$. >> $$Then,\ average\ speed,\ V_{avg} = \frac{(S1+S2+S3)}{(\frac{S1}{V1}+\frac{S2}{V2}+\frac{S3}{V3})}$$ >> $$t_1 = \frac{S_1}{V_1}, t_2 = \frac{S_2}{V_2}, t_3 = \frac{S_3}{V_3}$$ >5. A boy goes from home to school with speed $v1$ and come back to home with speed $v2$. Here distance covered by the boy is same time taken by the boy, from home to school, >> $$t_1 = \frac{S}{V_1}$$ >> and time taken by the boy, from school to home, >> $$t_2=\frac{S}{V_2}$$ >>> $$Then,\ average\ speed,\ V_av = \frac{S+S}{t₁+t₂} = \frac{2s}{(\frac{S}{V_1}+\frac{S}{V_2})} ⇒ V_av = \frac{2V_1V_2}{V_1+V_2}$$ >6. If an object covered $\frac{1}{3} rd$ distance with speed u, next $\frac{1}{3} rd$ with speed v and last $\frac{1}{3} rd$ distance with speed w then, >>> $$V_avg = \frac{3uvw}{uv + vw + wu}$$ #### Velocity and Acceleration 1. **Velocity:** It is define as the rate of change of displacement. > $$Therefore,\ velocity = \frac{displacement}{time} \text{ or it is the distance travelled in unit time in a given direction.}$$ >> $$velocity = \frac{distance\ travelled\ in\ a\ given\ direction}{time\ taken}$$ >>> S.I. unit of velocity is m/s. It is a vector quantity. (Magnitude of the velocity is known as speed) - **Types of Velocity** >1. **Uniform Velocity (Constant Velocity):** If a body covers equal distance in equal intervals of time in a given direction then it is said to be moving with constant velocity. >2. **Non-Uniform Velocity:** When a body does not cover equal distances in equal intervals of time, in a given direction (in this case speed is not constant), then it is known as non uniform velocity. >>> In uniform circular motion speed is constant but velocity is not constant. >3. **Average Velocity:** It is defined as the ratio of total displacement to the total time taken for this displacement. It is denoted by $V_{av}$ or v. It is a vector quantity. >>> $$Average\ velocity = \frac{Total\ displacement}{Total\ time}$$ >>>> i.e. $$v = \frac{s}{t}$$ >>>>> It is a vector quantity, and its direction is in the direction of displacement. - **Difference between Speed and Velocity** | S.No. | Speed | Velocity | |-------|-------|----------| | 1. | It is rate of change of position of an object. | It is rate of change of position of an object in a specific direction. | | 2. | Speed = $\frac{distance\ travelled}{time}$ | Velocity = $\frac{displacement}{time}$ | | 3. | It is a scalar quantity. | It is a vector quantity | | 4. | Speed will always be positive | It will be positive or negative depending on the direction of motion | | 5. | For moving body, it will never be zero | It may be zero | 2. **Acceleration:** Mostly the velocity of a moving object changes either in magnitude or in direction or in both when the object moves. The body is then said to have acceleration. So it is the rate of change of velocity > i.e. change in velocity in unit time is called acceleration. It is a vector quantity and Its S.I unit is $m/s²$ and c.g.s unit is $cm/s²$. >>> $$Acceleration = \frac{change\ in\ velocity}{time}$$ >>>> $$a = \frac{v-u}{t} = \frac{final\ velocity - initial\ velocity}{time\ taken}$$ - **Types of Acceleration** >1. **Positive acceleration:** If the velocity of an object increases with respect to time in the same direction, the object has a positive acceleration. >2. **Negative acceleration (retardation):** If the velocity of a body decreases with respect to time in the same direction, the body has a negative acceleration or it is said to be retarding. >> Eg. A train slows down, then its acceleration will be negative. #### Graphical Interpretation of some Quantities 1. **Position vs Time Graph:** >1. **Zero Velocity:** As position of particle is fixed at all the time, so the body is at rest. >> slope x-t graph gives velocity >>> Slope = tan 0 = tan 0° = 0 >>>> Velocity of particle is zero >2. **Uniform Velocity** >> Here tan 0 is constant >>> Slope is constant. >>>> so, velocity of particle is constant. >3. **Non uniform velocity (increasing with time)** >> In this case; >>> As time is o increasing is also increasing. >>>> Slope = tan 0 is also increasing Hence, velocity of particle is increasing. >4. **Non uniform velocity (decreasing with time)** >> In this case; >>> As time increases, 0 decreases. >>>> Slope = tan 0 also decreases. >>>>> Hence, velocity of particle is decreasing. 2. **Velocity vs Time Graph:** >1. **Zero acceleration** >> Velocity is constant. >>> Slope of v-t graph gives acceleration >>>> Slope = tan 0 = 0 >>>>> Hence, acceleration is zero. >>>>>> Area under v-t graph and time axis gives displacement. >2. **Uniform acceleration** >> tan 0 is constant. >>> Hence, it shows constant acceleration. >3. **Uniform retardation** >> Since 0 > 90° >>> tan 0 is constant and negative. >>>> Hence, it shows constant retardation. 3.**Acceleration vs Time Graph:** >1. **Constant acceleration** >> Hence, acceleration is constant. >>> area under a-t graph and time axis given change in velocity. #### Motion with uniform velocity Consider a particle moving along x-axis with uniform velocity u starting from the point $x = x_1$ at $t = 0$. Equations of x, v, a are : $x (t) = x_1 + ut$ ; $v (t) = u$ ; $a (t) = 0$ - x-t graph is a straight line of slope u through $x_1$. - as velocity is constant, v-t graph is a horizontal line. - a-t graph coincides with time axis because $a = 0$ at all time instants. #### Uniformly Accelerated Motion and Equations of Motion: If a particle is accelerated with constant acceleration in an interval of time, then the motion is termed as uniformly accelerated motion in that interval of time. There are three equation of uniformly accelerated motion. They show the relation between initial velocity u, final velocity v, acceleration a, time t and displacements. ##### Derivation of equations of Motion: 1. **$1$st equation of motion:** Consider a body moving with initial velocity u and its velocity changes from u to v in time t. Then > $$acceleration=\frac{Final\ velocity - Initial\ velocity}{time\ taken}$$ >>> $$a=\frac{v-u}{t}$$ >>>> $$So\ at=v-u\ and\ v=u+at$$ >>>>> $$1st\ equation\ of\ motion: v = u + at$$ 2. **$2$nd Equation of Motion:** We know > Distance covered = (Average velocity) × (Time) >>> or $$S = \frac{u + v}{2} t$$ >>>> But $$v = u + at$$ >>>>> Substituting the value of v in the equation above, we have >>>>>> $$s = \frac{u+(u+at)}{2} t=\frac{2u+at}{2} t$$ >>>>>> $$or\ s = 2ut + \frac{at^2}{2} =(u+\frac{at}{2}) t$$ >>>>>> $$or\ s= ut + \frac{1}{2}at^2$$ >>>>>>> $$2nd\ equation\ of\ motion: s = ut + \frac{1}{2}at^2 $$ 3. **$3$rd Equation of Motion:** We know that > $$v = u + at$$ >>> $$or\ t=\frac{v-u}{a}$$ >>>> Distance travelled = (Average velocity) × (time) >>>>> $$or\ S = \frac{u+v}{2} t = \frac{v+u}{2} \frac{v-u}{a}$$ >>>>>> $$or\ S = \frac{v^2-u^2}{2a} \text{ or } v^2 – u² = 2 as$$ >>>>>>> $$3rd\ equation\ of\ motion: v^2 – u² = 2 as$$ 4. **Distance travelled in nth second:** Distance travelled in n sec = Distance travelled in (n - 1) sec. > $$So, S_{nth} = S_n -S_{(n-1)}$$ >> $$=\frac{1}{2}[un+an^2] - [\frac{1}{2}(u(n-1)+a(n-1)^2)]$$ >>> [Putting t = n and t = (n-1) respectively in equation (ii)] >>>> $$=\frac{1}{2}un+\frac{1}{2}an^2 - \frac{1}{2}un+\frac{1}{2}u - \frac{1}{2}a(n^2-2n+1)$$ >>>>> $$=\frac{1}{2}u + (2n-1)\frac{1}{2}a$$ >>>>>> $$We\ have,\ S_{nth} = u + (2n-1)\frac{1}{2}a$$ ##### Mind it: - The magnitude of instantaneous velocity and instantaneous speed are equal. - Instantaneous velocity is always tangential to the path. ##### Graphical derivation of equations of motion: 1. **First Equation:** > $$v = u + at$$ > It can be derived from v - t graph, as shown in figure >> From line PQ, the slope of the line = acceleration >>> $$a = \frac{QR}{RP}=\frac{SP}{RP}$$ >>>> $$SP=v-u$$ >>>>> $$and\ RP = t\ So, a = \frac{v-u}{t}$$ >>>>>> $$or\ v = u + at$$ 2. **Second Equation:** > $$S = ut + \frac{1}{2} at^2$$ > It can also be derived from v – t graph as shown in figure. >> From relation, >>> Distance covered = Area under v t graph >>>> $$s = Area\ of\ trapezium\ OPQS = Area\ of\ rectangle\ OPRS + Area\ of\ triangle\ PQR$$ >>>>> $$= OP × PR + \frac{1}{2} RQ × PR$$ >>>>>> Putting values, >>>>>>> $$s= uxt + \frac{1}{2}(v-u)xt\ (:: RQ = v -u & PR = OS = t)$$ >>>>>>> $$= uxt + \frac{1}{2}at x t\ (.v-u = at)$$ >>>>>>> $$or\ s= ut + \frac{1}{2} at^2$$ 3. **Third Equation:** > $$v^2 = u^2 + 2aS$$ > From above graph OP = u, SQ = v, OP + SQ = u + v >> $$a = \frac{QR}{PR}$$ >>> $$Or\ PR = \frac{QR}{a} = \frac{v-u}{a}$$ >>>> S = Area of trapezium OPQS = $\frac{OP + SQ}{2}$ × PR >>>>> On putting the values, >>>>>> $$S = \frac{u + v}{2} \frac{v-u}{a}$$ >>>>>> $$or\ S = \frac{v^2-u^2}{2a}$$ >>>>>>> $$or\ v^2 = u^2 + 2as$$ ##### Equations of motion for freely falling object: Since the freely falling bodies fall with uniformly accelerated motion, the three equations of motion derived earlier for bodies under uniform acceleration can be applied to the motion of freely falling bodies. For freely falling bodies, the acceleration due to gravity is 'g', so we replace the acceleration 'a' of the equations by 'g' and since the vertical distance of the freely falling bodies is known as height 'h', we replace the distance 's' in our equations by the height 'h'. This gives us the following modified equations for the motion of freely falling bodies. 1. **Motion under gravity (uniformly accelerated motion):** > The acceleration with which a body travels under gravity is called acceleration due to gravity 'g'. Its value is $9.8 m/s²$ ( or ≈ $10 m /s²$). If you have to take $g = 10 m/s²$ then it must be mentioned in the question otherwise take $g = 9.8 m/s²$. - If a body moves upwards (or thrown up) g is taken negative (i.e. motion is against gravitation of earth). So equation of motion becomes. > $$v = u-gt,$$ >>> $$s=ut-\frac{1}{2}gt²$$ >>>> $$v² = u²- 2 gh$$ - If a body travels downwards (towards earth) then g is taken + ve. So equations of motion becomes > $$v = u + gt$$ >>> $$s = ut + \frac{1}{2}gt²$$ >>>> $$v² = u² + 2 gh$$ - If a body is projected vertically upwards with certain velocity then it returns to the same point of projection with the same velocity in the opposite direction. - The time for upward motion is the same as for the downward motion. ##### Sign Conventions: 1. g is taken as positive when it is acting in the same direction as that of motion and g is taken as negative when it is opposing the motion. 2. Distance measured upward from the point of projection is taken as positive, while distance measured downward from the point of projection is taken as negative. 3. Velocity measured away from the surface of earth (i.e. in upward direction) is taken as positive, while velocity measured towards the surface of the earth is taken as negative. ##### To Solve numerical problems: 1 If a body is dropped from a height then its initial velocity u = 0 but has acceleration (acting). If a body starts from rest its initial velocity $u = 0$. 2. If a body comes to rest, its final velocity $v = 0$ or, if a body reaches the highest point after being thrown upwards its final velocity $v = 0$ but has acceleration (acting). 3. If a body moves with uniform velocity, its acceleration is zero i.e. $a = 0$. 4. Motion of a body is called free fall if only force acting on it is gravity (i.e. earth's attraction). #### Solved Examples - **Example 1:** A body starts from A and moves according to given figure. (body retraces the path after C then reaches to D) > The distance and displacement are as follows for different path. >>> | Path | Distance | Displacement | >>> |---|---|---| >>> | AB | 4m | 4m | >>> | ABC | 10m | 10m | >>> | ABCB | 16m | 4m | >>> | ABCA | 20m | 0m | >>> | ABCAD | 25m | -5m | - **Example 2:** There are $n$ steps each of dimension $1$, $b$ & $h$ if a man climbs $n$ steps what is his displacement and distance. > By Pythagorean theorem $AB = \sqrt{b^2 + h^2}$. >> Similarly for each step >>> Displacement = $\sqrt{b^2 +h^2}$ >>>> So that total displacement = $n × \sqrt{b^2 + h^2}$ >>>>> $⇒$ Distance = $n (h + b)$ - **Example 3:** When the average speed of an object is equal to the magnitude of its average velocity ? Give reason also. > As average speed = $\frac{total\ path\ length}{time\ interval}$ >> Displacement >>> also, average velocity = $\frac{time\ interval}$ >>> When an object moves along a straight line or in the same direction its total path length is equal to the magnitude of its displacement. Hence average speed is equal to the magnitude of its average velocity. - **Example 4:** A person moves in a circular path centered at O. He starts from A and reaches diametrically opposite point B. Then find: (i) distance between A and B (ii) displacement between A and B > (i) Distance = Length of actual circular path from A to B = Half the circumference >> i.e. Distance = $\dfrac{2\pi r}{2} = \pi r$ >>> as r = 1m ... Distance = πm > (ii) Displacement = 2r along west. = 2m along west - **Example 5:** A car is moving along x-axis. As shown in figure it moves from O to P in $18$ s and returns from $P$ to $Q$ in $6$ second. What is the average velocity and average speed of the car in going from (i) $O$ to $P$ and (ii) from $O$ to $P$ and back to $Q$. > (i) Average velocity = $\frac{Displacement}{time\ interval} = \frac{360m}{18} = 20\ ms^{-1}$ >>> Average speed = $\frac{path\ length}{time\ interval}= \frac{360m}{18} = 20\ ms^{-1}$ > (ii) From $O$ to $P$ and back to $Q$ >> Average velocity = $\frac{OQ}{18+6} = \frac{240m}{24} = 10\ ms^{-1}$ >> Average speed = $\frac{path\ length}{time\ interval} = \frac{OP+PQ}{18+6} = \frac{360+120}{24} = 20\ ms^{-1}$ - **Example 6:** A car covers the 1st half of the distance between two places at a speed of $40 km h^{-1}$ and the 2nd half with $60 km h^{-1}$. What is the average speed of the car? > Suppose the total distance covered is $2S$. >> Then time taken to cover the distance 'S' with speed $40 km/h$, >>> $t_1 = \dfrac{S}{40} h,$ >>>> Time taken to cover next distance S with speed $60 km/h$ >>>>> $t_1 = \dfrac{S}{60} h,$ >>>>>> Average speed = $\dfrac{2S}{t_1 + t_2 } = \dfrac{2S}{\dfrac{S}{40}+\dfrac{S}{60}} = 48km/h$ - **Example 7:** A car covers the 1st half of the distance between two places at a speed of $40 km h^{-1}$ and the 2nd half with $60 km h^{-1}$. What is the average speed of the car? > Suppose the total distance covered is $2S$. >> Then time taken to cover the distance 'S' with speed $40 km/h$, >>> $t_1 = \dfrac{S}{40} h$ >>>> Time taken to cover the next distance 'S' with speed $60 km/hrs$, >>>>> $t_2= \dfrac{S}{60} h$ >>>>>> $V_av =\dfrac{total\ distance}{total\ time}= \dfrac{S+S}{(\frac{S}{40}+\frac{S}{60})} = \dfrac{2S}{\frac{3S+2S}{120}} = \dfrac{2S}{5S} × 120 = V_av = 48\ km/h$ - **Example 8:** A non-stop bus goes from one station to another station with a speed of $54 km/h$, the same bus returns from the second station to the first station with a speed of $36 km/h$. Find the average speed of the bus for the entire journey. Suppose the distance between the stations is $S$. Time taken in reaching from one station to another station. > $t_1 = \dfrac{S}{54} h$ >> Time taken in returning back, >>> $t_2 = \dfrac{S}{36} h$ >>>> Total time $t = t_1 + t_2$ >>>>> $t= \dfrac{S}{54}+\dfrac{S}{36} = \dfrac{2S+3S}{108} h$ >>>>>> Average speed = $\dfrac{Total\ distance}{Total\ time} = \dfrac{2S}{\dfrac{5S}{108}}h = \dfrac{2S}{5} × 108 = 43.2 km/h$ - **Example 9:** A particle starts from a point $A$ and travels along the solid curve shown in figure. Find approximately the position $B$ of the particle such that the average velocity between the positions $A$ and $B$ has the same direction as the instantaneous velocity at $B$. > The given curve shows the path of the particle starting at $y = $ $4$ m. >>> Average velocity = $\frac{displacement}{time\ taken}$ >>>> where displacement is straight line distance between points Instantaneous velocity at any point is the tangent drawn to the curve at that point. >>>>> Now, as shown in the graph, line $AB$ shows displacement as well as the tangent to the given curve. Hence, point B is the point at which direction of $AB$ shows average as well as instantaneous velocity. - **Example 10:** Velocity-time curve for a body moving with constant acceleration is shown in the figure. Calculate the displacement travelled by the body in 10 sec. > Displacement = area under the velocity time curve along time axis >> = area AOB >>> Now AOB is a triangle with base = $10$ sec and height = $10$ m/sec. >>>> So area = $\dfrac{1}{2} × Base × height$ >>>>> = $\dfrac{1}{2} × 10\ sec × 10\ m/sec$ >>>>>> = $5×10=50 m$ - **Example 11:** A ball is thrown vertically upwards with a velocity of $10m/sec$. It strikes the ground after $2$ sec. Its velocity-time graph is as shown in figure below. Find the displacement travelled by the ball in $2$ second. > Displacement = area under velocity-time curve along time axis >> = area of triangle AOB + area of triangle BDC >>> = $\dfrac{1}{2}× OB × AO + \dfrac{1}{2}× BD × CD$ >>>> = $\dfrac{1}{2}× 1\ sec × 10\ m/sec + \dfrac{1}{2}× 1\ sec × (-10m/sec)$ >>>>> = $5m-5m = 0m$ - **Example 12:** A stone drops from the edge of a roof. It passes a window $2$ metre high in $0.1$ second. How far is the roof above the top of the window? > Let the distance between the top of the window and the roof be $h$. This problem can be solved in two stages. >> For the journey across the window i.e., from B to C >>> Let, Velocity at B = $u$ m/s >>>> Distance travelled, $s = 2$ m >>>>> Time taken, $t = 0.1s$ >>>>>> Acceleration, $a = g=9.8 m/s²$ >>>>>>> Using the relationship, >>>>>>>> $s = ut + \frac{1}{2}gt²$ >>>>>>>> $s = ut + \frac{1}{2}gt²$ >>>>>>>> $2 = u × 0.1 + \frac{1}{2}× 9.8 × (0.1)²$ >>>>>>>> $2 = 0.1 u + 4.9 × 10^{-2}$ >>>>>>>> $(2 − 0.049)$ >>>>>>>> $or, u = \dfrac{}{(0.1)} = 19.51 m/s$ >>>>>>>> The velocity of the stone at the top of the window is $19.51 m/s$. >>>>>>>> For journey from roof to the top of the window i.e., from A to B >>>>>>>> The velocity at the top of the window is the velocity of the stone at the end of falling through 'h'. So, for this part of the journey, >>>>>>>> Initial velocity, $u = 0 m/s$ >>>>>>>> Final velocity, $v = $ $19.51 m/s$ >>>>>>>> Acceleration due to gravity, $g = 9.8 m/s²$ >>>>>>>> Then by using the equation, $v² – u² = 2 gh$, one gets >>>>>>>> $(19.51)²-0 = 2 × 9.8 × h$ >>>>>>>> $h = \dfrac{(19.51)^2}{19.6} m = 19.42 m$ >>>>>>>> Thus, the roof is $19.42$ m from the upper end (top) of the window. - **Example 13:** A ball is thrown vertically upwards with a velocity 'u'. Calculate the velocity with which it falls to the earth again. > For a ball thrown vertically upwards, >> Initial velocity = u >>> Final velocity = $v=0$ >>>> For the vertically upward motion, the equation of motion is >>>>> $v=u-gt$ >>>>>> So, $0=u-gt$ >>>>>>> $or\ t = \dfrac{u}{g}… (i)$ >>>> For the return journey, when the body falls vertically downwards, the equation of motion is >>>>> $v=u+gt$ >>>>>> Since, $u = 0$ >>>>>>> Hence $v=0+gt$ >>>>>>>> $or\ t = \dfrac{v}{g}… (ii)$ >>>> From (i) and (ii), >>>>> Thus, the body falls back to the earth with the same velocity with which it was thrown vertically upwards. - **Example 14:** A car is moving at a speed of $50 km/h$ after two seconds it is moving at $60 km/h$. Calculate the acceleration of the car. > Here $u = 50\ km/h = 50 × \dfrac{5}{18} m/s = \dfrac{250}{18}

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