Question

(II) A 975-kg sports car accelerates from rest to 95 km/h in 6.4 s. What is the average power delivered by the engine?

Short answer

The obtained value of average power delivered by the engine is \(5.3 \times {10^4}{\rm{ W}}\).

Step-by-step solution

Determine the energy transformed by the car

In this problem,to determine the average power of the car engine, you have to evaluate the amount of energy transformed by the engine. This energy could be divided bythe time consumed.

Given data:

The mass of the car is\(m = 975\;{\rm{kg}}\).

The final velocity of the car is\(v = 95\;{\rm{km/h}}\).

The final velocity of the car is\(u = 0\;{\rm{km/h}}\).

The time is\(t = 6.4\;{\rm{s}}\).

The relation to calculate energy transformed is given by:

\(\begin{aligned}E &= {E_{\rm{f}}} - {E_{\rm{i}}}\\E &= \left( {\frac{1}{2}m{v^2}} \right) - \left( {\frac{1}{2}m{u^2}} \right)\end{aligned}\)

Here, \({E_{\rm{i}}}\) is the initial kinetic energy, \({E_{\rm{f}}}\) is the final kinetic energy, and u is the initial velocity of the car whose value is zero.

On plugging the values in the above relation, you get:

\(\begin{aligned}E &= \left( {\frac{1}{2}m{v^2}} \right) - \left( {\frac{1}{2}m{{\left( 0 \right)}^2}} \right)\\E &= \left( {\frac{1}{2}\left( {975\;{\rm{kg}}} \right){{\left( {95\;{\rm{km/h}} \times \frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}} \right)}^2}} \right)\\E &= 339482.06\;{\rm{J}}\end{aligned}\)

Estimate the average power delivered by the engine

The relation of power is given by:

\(P = \frac{E}{t}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}P &= \left( {\frac{{339482.06\;{\rm{J}}}}{{6.4\;{\rm{s}}}}} \right)\\P &= 5.3 \times {10^4}{\rm{ W}}\end{aligned}\)

Thus, \(P = 5.3 \times {10^4}{\rm{ W}}\) is the power delivered by the engine.