Question

Constant power P is delivered to a car of mass m by its engine. Show that if air resistance can be ignored, the distance covered in a time t by the car, starting from rest, is given by s = ( 8P 9m ) 1/2 t 3/2 .

Answer 1

Answer:

Check Explanation.

Explanation:

At rest, the car had zero kinetic energy, that is K(rest) = 0. The kinetic energy of the car when the car moved, k(moved) can be calculated by using the work- energy relationship that is;

Work, W= k(moved) - k(rest).

k(moved) = work, W + k(rest).

[Recall that k(rest) = 0]. Therefore, k(moved) = work, W.

{Two Important equations to note are (1). Work, W=( 1/2) mass,m × (speed, V)^2 and (2). Power, P = work,W/ time,t}.

Hence, k(moved) = ( 1/2) mass,m × (speed, V)^2 ---------------------------(1).

Since, power, P= W/t -----------------(2).

Then, equation (2) becomes;

Power, P = m × v^2/ 2 × t.

Making, speed, v the subject of the formula, we have;

v = [(2 × P × t)/ m]^ 1/2. ---------------(3).

Another thing we have to remember is the formula for speed which is the change in distance with time that is; ds/ dt.

Therefore, ds/ dt = [(2 × P × t)/ m]^ 1/2.

ds/dt = [(2 × P)/ m]^ 1/2 × t^ 1/2 ------(4).

Then, the integration of the equation (4) above will give us;

s = ( 8P 9m )^ 1/2 × t^ 3/2 .

(Check attachment for more).

Answer 2

Answer:

Please look in the explanation section

Explanation:

Using the equation of kinetic energy:

$E_{k} =\frac{1}{2} mv^{2}$

Where

m is the mass, v is the velocity. The equation of power is:

$P=\frac{W}{t}$

Where W is the work and t is the time. Initially, the work is equal to the kinetic energy:

$W=E_{k-initial} \\P=\frac{E_{k-initial}}{t} \\P=\frac{mv^{2} }{2t}$

Clearing v:

$v=\sqrt{\frac{2tP}{m} }$

The velocity is:

$v=\frac{ds}{dt}$

Replacing:

$\frac{ds}{dt} =\sqrt{\frac{2tP}{m} }\\ds=\sqrt{\frac{2tP}{m} }dt$

Integrating:

$s=\sqrt{\frac{2P}{m} } (\frac{2t^{3/2} }{3} )=\sqrt{\frac{8P}{9m} } t^{3/2}$