A ball moves on a smooth floor in a straight line with a uniform velocity $10 m / s$ for $6 s$ . At $t = 6 s$ , the ball hits a wall and come back along the same line to the starting point with same speed. Draw the velocity time graph and use to it find the total distance travelled by the ball and its displacement.

Open in App

Solution

This is the required v - t graph
So distance will be equal for both forward and backward position

$d = v \times t$
$d = 10 \times 6$
so total distance is D=2d=120m

and
displacement=0
because initial and final position are same.

Suggest Corrections

Similar questions

Q. A body starting from rests accelerates uniformly along a straight line at the rate of

10 m s^{- 2}

for

5

seconds. It moves for

2

second with uniform velocity of

50 m s^{- 1}

. Then it retards uniformly and comes to rest in

3 s

. Draw a velocity-time graph of the body and find the total distance traveled by the body.

In a snooker match ball A is hit and it is moving with speed 4 m/s in the x-direction. It goes and hits balls B and C such that after collision ball B moves with velocity 4 m/s at angle $\circ 60$ to x-direction and A comes to rest. Find the velocity of ball C?

Q. A ball moving with a speed of 30 m/s hits the wall and bounces back with same speed. The contact time between the ball and the wall is 0.1 s. Calculate the force exerted by the wall on the ball if mass of the ball is 0.25kg.

Q. A ball of mass

10 g

moving perpendicular to the plane of the wall strikes it and rebounds in the same line with the same velocity. lithe impulse experienced by the wall is

0.54

N-s, the velocity of the ball is

Q. A particle is projected vertically upwards with velocity

40 m / s

. Find displacement and distance travelled by the particle in

6 s

Take

g = 10 m / s^{2}

Join BYJU'S Learning Program

This is the required v - t graph So distance will be equal for both forward and backward position d=v×t d=10×6 so total distance is D=2d=120mand displacement=0 because initial and final position are same.

This is the required v - t graph
So distance will be equal for both forward and backward position

$d = v \times t$
$d = 10 \times 6$
so total distance is D=2d=120m

and
displacement=0
because initial and final position are same.