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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. If the driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?

Respuesta :

Answer:

For 24 seconds force exerted is 5092 N towards opposite direction of motion of bus.

For 3.90 seconds force exerted is 31333 N towards opposite direction of motion of bus.

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 20 m/s

     Final velocity, v = 0 m/s    

Case 1:-

     Time, t = 24 s

     Substituting

                      v = u + at  

                      0 = 20 + a x 24

                      a = -0.8333 m/s²

     Force = Mass x Acceleration = 6110 x -0.8333 = -5092 N

     Force exerted is 5092 N towards opposite direction of motion of bus.

Case 2:-

     Time, t = 3.90 s

     Substituting

                      v = u + at  

                      0 = 20 + a x 3.90

                      a = -5.13 m/s²

     Force = Mass x Acceleration = 6110 x -5.13 = -31333 N

     Force exerted is 31333 N towards opposite direction of motion of bus.

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