Question

A 1500 kgcar starts from rest on a horizontal road and gains a speed of 72 km/hin 30 s. (a) What is its kinetic energy at the end of the 30 s? (b) What is the average power required of the car during the 30 sinterval? (c) What is the instantaneous power at the end of the 30 sinterval, assuming that the acceleration is constant?

Short answer

1. The kinetic energy at 30 s is$3.0\times {10}^{5}J$.
2. The average power required during 30 s is found to be $1.0\times {10}^{4}W$.
3. Instantaneous power at the end of 30 s will be $2.0\times {10}^{4}W$.

Step-by-step solution

The given data

The mass of the car is, m=1500 kg

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

The time of travel by car, t=30 s

Understanding the concept of kinematics

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)

a) Calculation of kinetic energy at the 30s

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

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

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

b) Calculation of the average power required at the 30s

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:$K{E}_{final}=3.0\times {10}^{5}J$

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

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

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

c) Calculation of the required instantaneous power

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:

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

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:

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

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