Asked by Ellen
A 1000Kg car is traveling at 40 m/s.
a.) if the car stops in 20 seconds, what is the deceleration rate?
b.) what friction force will stop the car in 20 seconds?
a.) if the car stops in 20 seconds, what is the deceleration rate?
b.) what friction force will stop the car in 20 seconds?
Answers
Answered by
Damon
acceleration = (final velocity - initial velocity)/change in time
a = (0 - 40) m/s /20 s
a = - 2 m/s^2
F = m a
F = 1000(-2) = -2000 Newtons
a = (0 - 40) m/s /20 s
a = - 2 m/s^2
F = m a
F = 1000(-2) = -2000 Newtons
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