As the driver steps on the gas pedal, a car of mass 1 140 kg accelerates from re
ID: 1454444 • Letter: A
Question
As the driver steps on the gas pedal, a car of mass 1 140 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 = 1.17 t - 0.210 t^2 + 0.240 t^3 (a) What is the change in kinetic energy of the car during the interval from t = 0 to t = 2.60 s? J (b) What is the minimum average power output of the engine over this time interval? W (c) Why is the value in part (b) described as the minimum value?Explanation / Answer
m = 1140 Kg
a = 1.17 t - 0.210 t^2 + 0.240 t^3
K.E = 1/2*m*Vf^2 - 1/2*m*vi^2
As it starts from rest, Vi = 0
Vf = a.dt
Vf = (1.17 t - 0.210 t^2 + 0.240 t^3).dt
Vf = 1.17/2*t^2 - 1/3 * 0.210 t^3 + 0.240/4 * t^4
At t = 2.60 ,
Vf = 5.47 m/s
K.E = 1/2*1140 * 5.47^2
K.E = 17054.9 J
(B)
Power, p = K.E/time
P = 17054.9/2.6 W
P = 6559.5 W
(C)
Because Maximum will be a Very big Value.
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