Question

Example 10. A body moves with a velocity of 2 m/s for 5 s, then its velocity increases uniformly to 10 m/s in next 5 s. Thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after 5 s.
(i) Plot a velocity-time graph for the motion of the body.
(ii) From the graph, find the total distance covered by the body after 2 s and 12 s.​

Answer 1

Answer:

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

Explanation:

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