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Q 3.2 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

(e) (A/B) overtakes (B/A) on the road (once/twice)

B starts after A at a higher speed and B overtakes A on the road once.

Q 3..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 ;

(d) A and B reach home at the (same/different) time

The time(x-coordinate) is different for both A and B when their position(y-coordinate) is equal to that of their home. Therefore A and B reach home at a different time.

Q 3.2  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

(c) (A/B) walks faster than (B/A)

The velocity of a particle is equal to the slope of its position-time (x-t) graph. Since the graph of B is steeper B walks faster than A.

Q 3.2  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

(b) (A/B) starts from the school earlier than (B/A)

From the graph, we can see that the position of A starts changing at time t=0 whereas in case of B  starts changing at some finite time and therefore A starts from the school earlier than B.

Q 3.2 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)

It is clear from the graph that A Lives closer to the school than B.

Q 3.28  The velocity-time graph of a particle in one-dimensional motion is shown in the figure:

Which of the following formulae are correct for describing the motion of the particle over the time-interval tto t2 :

(a)         $x(t_{2})=x(t_{1})+v(t_{2})(t_{2}-t_{1})+\frac{1}{2}a(t_{2}-t_{1})^{2}$

(b)        $v(t_{2})= v(t_{1})+a (t_{2}-t_{1})$

(c)     $v_{average }=\frac{\left [ x(t_{2})-v(t_{2}) \right ]}{(t_{2}-t_{1})}$

(d)      $a_{average }=\frac{\left [ v(t_{2})-v(t_{2}) \right ]}{(t_{2}-t_{1})}$

(e)         $x(t_{2})=x(t_{1})+v_{average}(t_{2}-t_{1})+\left ( \frac{1}{2} \right )a_{average}(t_{2}-t_{1})^{2}$

(f)       $x(t_{2})-x(t_{1})$  =area under the v-t curve bounded by the time axis and the dotted line shown.

Only the formulae given in (c), (d) and (f) are correct for describing the motion of the particle over the time-interval t1 to t2. The formulae given in (a), (b) and (e) are incorrect as from the slope of the graph we can see that the particle is not moving with constant acceleration.

Q 3.27  The speed-time graph of a particle moving along a fixed direction is shown in Figure. Obtain the distance traversed by the particle between

(b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

As the speed is increasing in the time interval t = 0 s to t = 5 s the acceleration is positive and can be given by Speed at t = 2 s is Speed at t = 5 s is v1 = 15 m s-1  t1 = 5 - 2 = 3 s  Distance travelled in interval t = 2 s to t = 5 s is s1 Acceleration is negative after t = 5 s but has the same magnitude a2 = -2.4 m s-2 Speed at t =  5 s is u2 = 12 m s-1 t2 = 6 - 5 = 1 s Distance...

Q 3.27  The speed-time graph of a particle moving along a fixed direction is shown in Figure. Obtain the distance traversed by the particle between

(a) t = 0 s to 10 s

What is the average speed of the particle over the intervals in (a) and (b)?

The distance traversed by the particle equals the area under the speed time graph The area under the curve is The particle has travelled a distance of 60 m from t=0 s to t=10 s. The average speed over this interval is

Q 3.26 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s-1 and 30 m s-1. Verify that the graph shown in Figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s-2. Give the equations for the linear and curved parts of the plot.

As both the stones are being accelerated due to gravity its effect will come on the relative motion only when one of them reaches the ground. Till that point of time, the relative velocity of the second stone would remain the same with respect to the first stone. Let us consider the upward direction to be positive. V1 = 15 m s-1 V2 = 30 m s-1 Vrel = V2 -V1 = 30 - 15 = 15 m s-1 Initial velocity...

Q 3.25 On a long horizontally moving belt (Fig. ), a child runs to and fro with a speed 9 km h-1  (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1. For an observer on a stationary platform outside, what is the

(c) time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?

The distance between the parents is s = 50 m The relative velocity of the child with respect to both his parents remains the same as the parents are also standing on the moving belt. v = 9 km h-1 = 2.5 m s-1 Time taken by the child in (a) and (b) is t = s/v = 20 s. As both, the parents are also standing on the belt as well the speed of the child would appear to be 9 km h-1 to both the parents...

Q 3.25 On a long horizontally moving belt (Fig.), a child runs to and fro with a speed 9 km h-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1. For an observer on a stationary platform outside, what is the

(b) speed of the child running opposite to the direction of motion of the belt?

(b) Speed of the child with respect to the stationary observer when the boy is running in the direction opposite to the motion of the belt = 9 - 4 = 5 km h-1

Q 3.25  On a long horizontally moving belt (Fig.), a child runs to and fro with a speed 9 km h-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt?

(a) Speed of the child when the boy is running in the direction of the motion of the belt = 9 + 4 = 13 km h-1

Q 3.24  A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s-1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

Let us consider the upward direction to be positive Initial velocity of the ball (u) = 49 m s-1 The speed of the ball will be the same when it reaches the boy's hand's but will be moving in a downward direction. Therefore final velocity (v) = -49 m s-1 Acceleration (a) = -9.8 m s-2 Using the first equation of motion we have In the second case, as the ball has been thrown after the lift has...

Q 3.23  A three-wheeler starts from rest, accelerates uniformly with 1 m s-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

Initial velocity u = 0 Acceleration, a = 1 ms-2 t = n seconds Let the total distance travelled in n seconds be Sn Similarly, total distance travelled in n - 1 second would be Sn-1 Distance travelled in nth second would be given as xn = Sn - Sn-1 As we can see the dependency of xn on n is linear we conclude the plot of the distance covered by the vehicle during the nth second versus n would...

Q 3.22     Figure  gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

The average acceleration greatest in magnitude in interval 2 as in this interval the magnitude of change in speed is the greatest. The average speed is greatest in interval 3.   v a Interval 1 Positive Positive Interval 2 Positive Negative Interval 3 Positive Zero In interval 1 speed increases, in interval 2 it decreases and in interval 3 it remains constant. At points, A, B, C...

Q 3.21 Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

The average speed is greatest in interval 3 and least in interval 2 as the average of the magnitude of the slope is maximum in 3 and minimum in 2. Average velocity is positive in interval 1 and interval 2 as the slope is positive over these intervals and average velocity is negative in interval 3 as the slope is negative over this interval.

Q 3.20 Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.

At t=0.3 s signs of position, velocity and acceleration are negative, negative and positive. At t=1.2 s signs of position, velocity and acceleration are positive, positive and negative. At t=-1.2 s signs of position, velocity and acceleration are negative, positive and positive. Note: The displacement of the particle as a function of time can be thought of as  where A is some positive real...

Q 3.19 Suggest a suitable physical situation for each of the following graphs.

(a)The particle is initially at rest. Then it starts moving with a constant velocity for some time and then its velocity changes instantaneously and it starts moving in the opposite direction, crosses the point where it was at rest initially and then comes to a halt. A similar physical situation arises when a bowler throws a ball towards the batsman, the ball travels towards the batsman with...

Q 3.18     A police van moving on a highway with a speed of 30 km h-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h-1. If the muzzle speed of the bullet is 150 m s-1, with what speed does the bullet hit the thief’s car ?

Muzzle speed of the bullet is VB=150 m s-1=540 km h-1 Speed of the Police van is VV=30 km h-1 Resultant Speed of the bullet is V=540+30=570 km h-1         Speed of the thief's car= VT=192 km h-1 The speed with which the bullet hits the thief's car=V-VT=570-192=378 km h-1 =105 m s-1

Q 3.16    Look at the graphs (a) to (d)  carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

(a) The given x-t graph cannot represent the one-dimensional motion of a particle as the particle cannot be at two positions at the same instant. (b) The given v-t graph cannot represent the one-dimensional motion of a particle as the particle cannot be travelling at two velocities at the same instant. (c) The given speed-time graph cannot represent the one-dimensional motion of a particle as...
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