Question

A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it travel before coming to a stop? (b) How long does it take to stop from 24.0 m/s?

Answer 1

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Step-by-step explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that
`v^2=u^2+2as`

`16^2=24^2+2* a* 50`

`a=-3.2m/sec^2`

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So
`v^2=u^2+2as`

`0^2=24^2-2* 3.2* s`

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration
`a=-3.2m/sec^2`

So time
`t=(v-u)/(a)=(0-24)/(-3.2)=7.5sec`