As the driver steps on the gas pedal, a car of mass 1 180 kg accelerates from re

Question

As the driver steps on the gas pedal, a car of mass 1 180 kg accelerates from rest. During the first few seconds of motion, the car's acceleration increases with time according to the expression below, where t is in seconds and a is in m/s^2. a = 1.18 t- 0.210 t^2 + 0.240 t^3 What is the change in kinetic energy of the car during the interval from t = 0 to t = 3.50 s? What is the minimum average output of the engine over this time interval? Why is the value in part (b) described as the minimum value?

Explanation / Answer

a = dv/dt

dv = integral a.dt from 0 to 3.5s

v = 1.18t^2/2 -0.210t^3/3 + 0.240t^4/4 from 0 to 3.5 s

v = 13.23 m/s

KE = 1/2mv^2

KE = 103269.4 J

part b )

P = U/dt = KE/dt

P = 29505.546 W

part c )

The value in (b) represents only energy that leaves the engine and is transformed to kinetic energy of the car

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