Question

A body at rest is given an initial uniform acceleration of 8m/s2 for 30s after which the acceleration is reduced to 5m/s2 for the next twenty seconds. The body maintains the speed attained for 60s, after which it is brought to rest in 20s.​

Answer 1

Step-by-step explanation:

v=u+at, where v,u,a and t are the final velocity, initial velocity, acceleration and time, respectively.

For t=0 s to t=30 s , a =8 m/s 2.

For t=30 s to t=50 s , a = 5 m/s 2.

For t=50 s to t=110 s , a = 0 m/s 2.

For t=110 s to t=130 s , a is such that the body comes to rest at the end of this period.

For t=0 s to t=30 s ,v=0+8t⇒v=8t m/s.

At t=30s,v=8×30=240 m/s.

For t=30 s to t=50 s ,v=240+5(t−30)⇒v=90+5t m/s

At t=50s,v=90+5×50=340 m/s.

For t=50 s to t=110 s ,v=340 m/s.

For t=110 s to t=130 s ,v=340+a(t−110).

The body comes to a halt at t=130 s ⇒0=340+a(130−110).

0=340+20a⇒a=−34020=−17 m/s2.

For t=110 s to t=130 s ,v=340−17(t−110)=2210−17t m/s.