**1** **In which of the following examples of motion, can the body be considered approximately a point object:$\\$ (a) a railway carriage moving without jerks between two stations.$\\$ (b) a monkey sitting on top of a man cycling smoothly on a circular track.$\\$ (c) a spinning cricket ball that turns sharply on hitting the ground.$\\$ (d) a tumbling beaker that has slipped off the edge of a table.**

(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.$\\$ (b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.$\\$ (c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.$\\$ (d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

**2** **The position-time $(x-t)$ graphs for two children $A$ and $B$ returning from their school $O$ to their homes $P$ and $Q$ respectively are shown in figure. Choose the correct entries in the brackets below;$\\$ (a) $(A/B)$ lives closer to the school than $(B/A)$$\\$ (b) $(A/B)$ starts from the school earlier than $(B/A)$$\\$ (c) $(A/B)$ walks faster than $(B/A)$$\\$ (d) $A$ and $B$ reach home at the (same/different) time$\\$ (e) $(A/B)$ overtakes $(B/A)$ on the road (once/twice).**

(a) As $OP < OQ, A$ lives closer to the school than $B.$$\\$ (b) For $x = 0, t = 0$ for $A$; while $t$ has some finite value for $B$. Therefore, $A$ starts from the school earlier than $B.$$\\$ (c) Since the velocity is equal to slope of $x-t$ graph in case of uniform motion and slope of $x-t$ graph for $B$ is greater that that for $A$ =, hence $B$ walks faster than $A.$$\\$ (d) It is clear from the given graph that both $A$ and $B$ reach their respective homes at the same time.$\\$ (e) $B$ moves later than $A$ and his/her speed is greater than that of $A.$ From the graph, it is clear that $B$ overtakes$A$ only once on the road.

**3** **A woman starts from her home at $9.00$ am, walks with a speed of $5 km /h ^{- 1}$ on a straight road up to her office $2.5 km$ away, stays at the office up to $5.00$ pm, and returns home by an auto with a speed of $25 km /h^{- 1 }$. Choose suitable scales and plot the $x-t$ graph of her motion**

Speed of the woman $= 5 km/h$$\\$ Distance between her office and home $= 2.5 km$$\\$ Time taken = Distance / Speed$\\$ $= \frac{2.5} {5 }= 0.5 h = 30 min$$\\$ It is given that she covers the same distance in the evening by an auto. Now, speed of the auto $= 25 km/h$$\\$ Time taken = Distance /Speed $\\$ $= \frac{2.5 }{ 25} = \frac{1}{ 10 }= 0.1 h = 6 min$$\\$ The suitable $x-t$ graph of the motion of the woman is shown in the given figure.

**4** **A drunkard walking in a narrow lane takes $5$ steps forward and $3$ steps backward, followed again by $5$ steps forward and $3$ steps backward, and so on. Each step is $1$ m long and requires $1$ s. Plot the $x-t$ graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit $13 m$ away from the start.**

Distance covered with $1$ step = $1 m$$\\$ Time taken = $1 s$$\\$ Time taken to move first $5 m$ forward = $5 s$$\\$ Time taken to move $3 m$ backward = $3 s$$\\$ Net distance covered =$ 5 - 3 = 2 m$$\\$ Net time taken to cover $2 m = 8 s$$\\$ Drunkard covers $2 m$ in $8 s.$$\\$ Drunkard covered $4 m$ in $16 s.$$\\$ Drunkard covered $6 m$ in $24 s.$$\\$ Drunkard covered $8 m$ in $32 s.$$\\$ In the next $5 s$, the drunkard will cover a distance of $5 m$ and a total distance of $13 m$ and falls into the pit.$\\$ Net time taken by the drunkard to cover $13 m = 32 + 5 = 37 s$$\\$ The $x-t$ graph of the drunkard’s motion can be shown as:

**5** **A jet airplane travelling at the speed of $500 km /h^{ -1}$ ejects its products of combustion at the speed of $1500 km /h^{-1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?**

Speed of the jet airplane, $= 500 km/h$$\\$ Relative speed of its products of combustion with respect to the plane,$\\$ Speed of its products of combustion with respect to the ground$\\$ Relative speed of its products of combustion with respect to the airplane,$\\$ The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

**6** **A car moving along a straight highway with speed of $126 km h^{-1}$ is brought to a stop within a distance of $200 m$. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?**

Initial velocity of the car,$\\$ Final velocity of the car,$\\$ Distance covered by the car before coming to rest,$\\$ Retardation produced in the car$\\$ From third equation of motion, a can be calculated as:$\\$ From first equation of motion, time (t) taken by the car to stop can be obtained as:$\\$ $v=u+at\\ t=(v-u)/a=(-35)/(-3.06)=11.44 s$

**7** **Two trains A and B of length $400 m$ each are moving on two parallel tracks with a uniform speed of $72 km h^{-1}$ in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by $1 m s^{ -2}$ . If after $50 s$, the guard of B just brushes past the driver of A, what was the original distance between them ?**

For train A:$\\$ Initial velocity, $u = 72 km/h = 20 m/s$$\\$ Time,$ t = 50 s$$\\$ Acceleration, $a _I = 0 $(Since it is moving with a uniform velocity)$\\$ From second equation of motion, distance ($s _I $) covered by train A can be obtained as:$\\$ $s=ut+(1/2)a_1t^2$$\\$ $=20 * 50 +0=1000 m$$\\$ For train B:$\\$ Initial velocity, $u = 72 km / h = 20 m / s$$\\$ Acceleration,$ a = 1 m / s^ 2$$\\$ Time, $t = 50 s$$\\$ From second equation of motion, distance $(s _{II} )$ covered by train A can be obtained as:$\\$ $s_{ II}= ut+( 1/ 2 ) at^ 2\\ =20 *50 +( 1/ 2 )* 1 *( 50 )^2=2250 m$$\\$ Length of both trains $= 2 × 400 m = 800 m$$\\$ Hence, the original distance between the driver of train A and the guard of train B is $2250 - 1000 - 800 = 450m.$

**8** **On a two-lane road, car A is travelling with a speed of $36 km h^{- 1}$ . Two cars B and C approach car A in opposite directions with a speed of $54 km h ^{- 1}$ each. At a certain instant, when the distance AB is equal to AC, both being $1 km,$ B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?**

Velocity of car A,$\\$ Velocity of car B,$\\$ Velocity of car C,$\\$ Relative velocity of car B with respect to car A,$\\$ $= 15 - 10 = 5 m/s$$\\$ Relative velocity of car C with respect to car A,$\\$ $= 15 + 10 = 25 m/s$$\\$ At a certain instance, both cars B and C are at the same distance from car A i.e.,$\\$ $s = 1 km = 1000 m$$\\$ Time taken (t) by car C to cover $1000 m = 1000 / 25 = 40 s$$\\$ Hence, to avoid an accident, car B must cover the same distance in a maximum of $40 s.$$\\$ From second equation of motion, minimum acceleration (a) produced by car B can be obtained as: $s = ut +( 1/ 2 )at ^2\\ 1000=5*40+(1/2)*a*(40)^2\\ a=1600/1600=1 ms^{-2}$

**9** **Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of $20 km h ^{-1}$ in the direction A to B notices that a bus goes past him every $18$ min in the direction of his motion, and every $6$ min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?**

Let V be the speed of the bus running between towns A and B.$\\$ Speed of the cyclist,$ v = 20 km / h$$\\$ Relative speed of the bus moving in the direction of the cyclist $=V - v =( V - 20 ) km / h$$\\$ The bus went past the cyclist every $18$ min i.e., $18 / 60 h$ (when he moves in the direction of the bus).$\\$ Distance covered by the bus =$ (V - 20) × 18 / 60 km .... (i)$$\\$ Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to $V × T / 60 ....(ii)$$\\$ Both equations (i) and (ii) are equal. $(V - 20) × 18 / 60 = VT / 60 ......(iii)$$\\$ Relative speed of the bus moving in the opposite direction of the cyclist $= (V + 20) km/h$$\\$ Time taken by the bus to go past the cyclist =$ 6 min = 6 / 60 h$$\\$ $\therefore (V + 20) × 6 / 60 = VT / 60 ....(iv)$$\\$ From equations (iii) and (iv), we get$\\$ $(V + 20) × 6 / 60 = (V - 20) × 18 / 60\\ V + 20 = 3V - 60\\ 2V = 80\\ V = 40 km/h$$\\$ Substituting the value of V in equation (iv), we get$\\$ $(40 + 20) × 6 / 60 = 40T / 60\\ T = 360 / 40 = 9 min$

**10** **A player throws a ball upwards with an initial speed of $29.4 m s ^{- 1}$ .$\\$ (a) What is the direction of acceleration during the upward motion of the ball ?$\\$ (b) What are the velocity and acceleration of the ball at the highest point of its motion ?$\\$ (c) Choose the $x = 0 m$ and $t = 0 s$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.$\\$ (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take $g = 9.8 m s ^{- 2}$ and neglect air resistance).$\\$**

(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.$\\$ (b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., $9.8 m / s^ 2 .$$\\$ (c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.$\\$ (d) Initial velocity of the ball,$ u = 29.4 m/s$$\\$ Final velocity of the ball, $v = 0$ (At maximum height, the velocity of the ball becomes zero) Acceleration, $a = - g =- 9.8 m / s ^2$$\\$ From third equation of motion, height (s) can be calculated as:$\\$ $v^2-u^2=2gs\\ s=(v^2-u^2)/2g\\ =((0)^2-(29.4)^2)/2*(-9.8)=3s$$\\$ Time of ascent = Time of descent$\\$ Hence, the total time taken by the ball to return to the player’s hands $= 3 + 3 = 6 s$