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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 :

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 :

[tex]Ft=mv\\\\F=\dfrac{mv}{t}\\\\F=\dfrac{6110\times 20}{24}=5091.67\ N[/tex]

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 :          

[tex]F't=mv\\\\F'=\dfrac{mv}{t'}\\\\F'=\dfrac{6110\times 20}{3.9}=31333.34\ N[/tex]

Hence, this is the required solution.                            

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