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/s2. a = 1.14 t - 0.210 t2 + 0.240 t3 (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?

Answer 1

Answer:

(a) $KE=16405.215 J$

(b) P = 6309.6981 W

(c) Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

Explanation:

Given:

mass of car, m = 1140 kg

expression of acceleration, $a=1.14t-0.210t^2+0.240t^3$

where "t" is time in seconds

initial time, $t_i=0 s$

final time, $t_f=2.6 s$

(a)

We know,

$\frac{dv}{dt} =a$

$dv=a.dt$

$v=\int\limits^{2.6}_0 {1.14t-0.210t^2+0.240t^3} \, dt$

$v=5.3648 m.s^{-1}$

Kinetic Energy

∴$KE= \frac{1}{2} m.v^2$

$KE=\frac{1}{2}\times 1140\times 5.3648^2$

$KE=16405.215 J$

(b)

We know,

Power

$P= \frac{\Delta KE}{\Delta t}$

$P=\frac{16405.215}{2.6}$

P = 6309.6981 W

(c)

Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.