A 6110-kg bus traveling at 20.0m/s can be stopped in 24.0 s by gently applying the brakes. What is the average force exerted on the bus in this stop?
A. 1.222 x 10^3 N
B. 2.55 x 10^3 N
C. 5.09 x 10^3 N
D. 7.33 x 10^3 N
A. 1.222 x 10^3 N
B. 2.55 x 10^3 N
C. 5.09 x 10^3 N
D. 7.33 x 10^3 N
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Fnet = ma
find acceleration:
Vf = 0 m/s
Vi = 20.0 m/s
t = 24.0 s
Vf = Vi + at
0 = (20.0 m/s) + a(24.0s)
-20.0 m/s = a(24.0s)
a = -0.833 m/s^2
Fnet = ma
= (6110 kg)(-0.833)
= 5091.66667
= 5.09 x 10^3
find acceleration:
Vf = 0 m/s
Vi = 20.0 m/s
t = 24.0 s
Vf = Vi + at
0 = (20.0 m/s) + a(24.0s)
-20.0 m/s = a(24.0s)
a = -0.833 m/s^2
Fnet = ma
= (6110 kg)(-0.833)
= 5091.66667
= 5.09 x 10^3