Question

A 6110 kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. Ifthe driver slams on the brakes, the stops in 3.90 s. What is the average for exerted on the bus in both these stops

Answer 1

Step-by-step explanation:

Given that,

Mass of bus, m = 6110 kg

Speed of bus, v = 20 m/s

The bus will stop in 24 s gently applying the brakes, t = 24 s

The average force exerted on the bus is given by :

`Ft=mv\\\\F=(mv)/(t)\\\\F=(6110* 20)/(24)=5091.67\ N`

If the driver slams on the brakes, the stops in 3.90 s, t' = 3.9 s

The average force exerted on the bus at this time is given by :

`F't=mv\\\\F'=(mv)/(t')\\\\F'=(6110* 20)/(3.9)=31333.34\ N`

Hence, this is the required solution.