(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/s, neglecting friction?

(b) How long will this acceleration take if the car also climbs a 3.00-m high hill in the process?

The power or rating of a device is defined as the energy consumed by the appliances in a given time.

Mathematically,

\(P = \frac{W}{T}\)

Here, W is the work done and T is the time taken.

(a)

According to the work-energy theorem, the work done equals the change in the kinetic energy. Mathematically,

\({W_k} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\)

Here, m is the mass of the car\(\left( {m = 850{\rm{ kg}}} \right)\),\({v_f}\)is the final velocity\(\left( {{v_f} = 15.0{\rm{ m}}/{\rm{s}}} \right)\), and\({v_i}\)is the initial velocity\(\left( {{v_i} = 0} \right)\).

Putting all known values,

\(\begin{aligned}{W_k} &= \frac{1}{2} \times \left( {850{\rm{ kg}}} \right) \times {\left( {15.0{\rm{ m}}/{\rm{s}}} \right)^2} - \frac{1}{2} \times \left( {850{\rm{ kg}}} \right) \times {\left( 0 \right)^2}\\ &= 95625{\rm{ J}}\end{aligned}\)

Rearranging equation (1.1) in order to get an expression for time,

\(T = \frac{{{W_k}}}{P}\)

Here,\({W_k}\)is the work done by kinetic energy, and P is the power\(\left( {P = 40.0{\rm{ hp}}} \right)\).

Putting all known values,

\(\begin{aligned}T &= \frac{{95625{\rm{ J}}}}{{40.0{\rm{ hp}}}}\\ &= \frac{{95625{\rm{ J}}}}{{40.0{\rm{ hp}} \times \left( {\frac{{746{\rm{ W}}}}{{1{\rm{ hp}}}}} \right)}}\\ &= 3.2{\rm{ s}}\end{aligned}\)

Therefore, the car will take 3.2 s to reach a speed of 15 m/s.

(b)

The work done against gravity is,

\(\begin{aligned}{W_p} &= {E_p}\\ &= mgh\end{aligned}\)

Here, m is the mass of the car \(\left( {m = 850{\rm{ kg}}} \right)\), g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and h is the height of the hill \(\left( {h = 3.00{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned}{W_P} &= \left( {850{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3.00{\rm{ m}}} \right)\\ &= 24990{\rm{ J}}\end{aligned}\)

As a result, the total work done is,

\(W = {W_k} + {W_p}\)

Putting all known values,

\(\begin{aligned}W &= \left( {95625{\rm{ J}}} \right) + \left( {24990{\rm{ J}}} \right)\\ &= 120615{\rm{ J}}\end{aligned}\)

The time taken to climb the hill is,

\(T = \frac{W}{P}\)

Putting all known values,

\(\begin{aligned}T &= \frac{{120615{\rm{ J}}}}{{40.0{\rm{ hp}}}}\\ &= \frac{{120615{\rm{ J}}}}{{40.0{\rm{ hp}} \times \left( {\frac{{746{\rm{ W}}}}{{1{\rm{ hp}}}}} \right)}}\\ &= 4.0{\rm{ s}}\end{aligned}\)

Therefore, the car will take 4 s to climb a 3 m high hill in the process.