Question

19. An object moves with a velocity of 2 m/s for 5 s, then its velocity Meases un
in next 5 s. Therefore, its velocity begins to decrease at a uniform rate until it come
5 s.
(a) Plot a velocity-time graph for the motion of the body.
(b) From the graph, find the total distance covered by the body after 5 s and 10 s.




plz help me ​

Answer 1

Explanation:

The body moves from A to B at velocity 2 m/s for 5 s, then it accelerates, from B to C, for 5 s and attains a velocity 10 m/s. Finally, it comes to rest in 10 s at a constant retardation, from C to D.

2.

A to B is uniform motion.

B to C and C to D is non-uniform motion.

3.

Distance travelled in 2 s

From the graph, the distance travelled in 2 s is = ut = 2 × 2 = 4 m (1)

Distance travelled in 12 s

Distance travelled in 5 s = ut = 2 × 5 = 10 m ..(2)

In the next 5 s the acceleration is, a = (v-u)/t = (10-2)/5 = 1.6 m/s 2

Distance travelled in this 5 s is, S = ut + ½ at 2

=> S = 2 × 5 + 0.5 × 1.6 × 5 2

=> S = 30 m (3)

After that the body starts slowing down. The acceleration now is = (0-10)/10 = -1 m/s 2 (negative sign indicates deceleration)

So, distance travelled in 2 s during the slowing down, S = ut + ½ at 2

=> S = 10 × 2 0.5 × 1 × 2 2

=> S = 18 m (4)

So, total distance covered in 12 s is = 10+30+18 = 58 m

Distance travelled in 20 s

Distance travelled in 20 s is equal to (2)+(3)+total distance travelled during the slowing down.

Total distance travelled during the slowing down can be found using,

v 2 = u 2 + 2aS

=> 0 = 10 2 2 × 1 × S

=> S = 50 m

So, total distance travelled in 20 s = 10+30+50 = 90 m