Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed-squared. For a certain car with a weight of 12,000 N, the total resistant force, Fis given by $\mathit{F}\mathbf{=}\mathbf{300}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{8}{\mathit{v}}^{\mathbf{2}}$, with Fin newton, and, v in meters per second. Calculate the power (in horsepower) required to accelerate the car at$\mathbf{0}\mathbf{.}\mathbf{92}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$when the speed is 80 km/h.

1. Weight of the car, W = 12,000N
2. The resistance force,$F=300+1.8v^2$
3. Acceleration of the car, $a=0.92m/s^2$
4. Speed of the car, v = 80km/h

We use Newton’s second law to find the force applied by the car. The given velocity is in km/h we can convert it into m/s. Using that force and the velocity, we can find power. Then we convert this power to horsepower.

Formulae:

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

The instantaneous power obtained by the body, P = Fv (ii)

The value of the velocity in m/s is given as:

$\begin{array}{l}v=\frac{80\mathrm{km}}{h}\times \frac{1000m}{1km}\times \frac{1h}{3600s}\\ =22.22m/s\end{array}$

Now, the mass of the car from given weight is given as:

$$\begin{gathered} m=\frac{W}{g} \\ =\frac{12000}{9.8} \\ =1224.48kg \end{gathered}$$

Now we have to find theapplied force by the car, this can be given as:

$F_{net}=F_a-F$

where, $F_a$is applied force and F is resistance force.

The above equation is given using equation (i) as follows:

$\begin{array}{l}ma=F_a-\left(300+1.8v^2\right)\\ F_a=(1224.48\times 0.92\}+\left(300+1.8\times 22.{22}^2\right)\\ =2315.2327N\end{array}$

So, the power value is given using equation (ii) as:

$$\begin{gathered} P=2315.2327\times 22.22 \\ =5.14\times {10}^{4}W \\ =5.14\times {10}^{4}W\times \frac{1hp}{746W}\left(\because 1hp=746W\right) \\ =68.96hp \\ \approx 69hp \end{gathered}$$

Hence, the value of the power is 69hp.