Question

a body moves with a velocity of 2m/s for 5s. then its velocity increases uniformly to 10m/s in next 5s. thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after 10s
(a) plot a velocity-time graph
(b) mark the portions of the graph to show when the motion of the body is uniform and when it is non-uniform
(c) from the graph, find the total distance moved by the body in initial 5s, in initial 12s and in the last 10s

Answer 1

$\huge\bigstar\underline\mathfrak\blue{Solution}$

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 + 1/2 at²

=> 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 + 1/2at²

=> 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

Answer 2

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 + 1/2 at²

=> 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 + 1/2at²

=> 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