91Ó°ÊÓ

The mass of the car is, m=1500 kg

The speed of the car is,$v=72km/h×\left(\frac{5m/s}{18km/h}\right)=20m/s$

The time of travel by car, t=30 s

We are given the mass of the car. The velocity of the car after 30 s is also given. With these, we can find the kinetic energy of the car. Using the change in the kinetic energy in the given interval of time, we can find the average power during 30 s. Using the same procedure, we can find instantaneous power at 30 s.

Formulae:

The force due to Newton’s second law, F=ma (1)

The kinetic energy of the body, $KE=\frac{1}{2}m{v}^2$ (2)

The average power of the body, $P_{avg}=\frac{∆KE}{∆t}$ (3)

The instantaneous power generated by the body, P=Fv (4)

The first equation of kinematic motion, $v_f=v_i+at$ (5)

Using equation (2), the kinetic energy of the body at t = 30 s is given as:

$$\begin{gathered} K{E}_{\left(t=30s\right)}=\frac{1}{2}×1500×{(20m/s)}^2 \\ =3.0×{10}^{5}kg.m^2/s^2\left(\frac{1J}{1kg.m^2/s^2}\right) \\ =3.0×{10}^{5}J \end{gathered}$$

Hence, the value of the energy is $3.0×{10}^5\mathrm{J}$.

Initially, the car starts from rest. Thus, the initial energy is given as:$K{E}_{initial}=0J$

And, from part (a), the final velocity is given as:$K{E}_{final}=3.0×{10}^{5}J$

Thus, the average power of the block is given using equation (iii) as given:

$$\begin{gathered} P_{avg}=\frac{\left(3.0×{10}^{5}J\right)-\left(0J\right)}{\left(30s\right)-\left(0s\right)} \\ =\frac{\left(3.0×{10}^{5}J\right)}{\left(30s\right)} \\ =1.0×{10}^{4}W \end{gathered}$$

Hence, the value of the average power is $1.0×{10}^{4}W$.

As the car starts to rest,$v_i=0m/s$

Thus the final velocity of the block is given using equation (5):$v_f=at$

Using equations (1) and the above value in equation (4), we can get that the instantaneous power required by the block is given as:

$$\begin{gathered} \mathrm{P}=ma\left(at\right) \\ =m{a}^{2}t…....………\left(6\right] \end{gathered}$$

Initially, the car starts from rest:$K{E}_{initial}=0J,t_{initial}=0s$

Thus, the average power of the body is given using equations (2) in equation (3) as:

$$\begin{gathered} P_{avg}=\frac{m{v}^2}{2t} \\ =\frac{m{a}^2{t}^2}{2t}(fromequation(5\left)\right) \\ =\frac{m{a}^{2}t}{2t}....................\left(6\right) \end{gathered}$$

Now, from equations (5) and (6), we can get the required instantaneous power as:

$$\begin{gathered} P=2×P_{avg} \\ =2×1.0×{10}^{4}W \\ =2.0×{10}^{4}W \end{gathered}$$

Hence, the value of the power is $2.0×{10}^{4}W$.