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

Answer:

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

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

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 $v^2=u^2+2as$

$16^2=24^2+2\times a\times 50$

$a=-3.2m/sec^2$

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

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

$0^2=24^2-2\times 3.2\times s$

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 $t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec$