A car of mass 1000kg can deliver constant mechanical power of 7460 W. If at a certain instant the car is moving with just half the maximum velocity it can attain against a resistance of 2000 N, find its acceleration at that instant  $\left(inm{s}^{-2}\right)$.  

Let the total resistance acting on the car be R.
At the instant when the car is moving with a speed v, the mechanical power and force developed by the engine be P and F respectively. Then
P = Fv and   (i)
F - R = ma  (ii)
From (i), it is clear that smaller the velocity v of the car greater is force F developed for a given power. From (ii), greater the F, greater the acceleration a.
$$\begin{gathered} \therefore P_{\max }=F_{\min }\times v_{\max } \\ But{F}_{\min }=R \\ Hence{v}_{\max }=\frac{P_{\max }}{R}=\frac{7460}{2000}=3.73m/s \end{gathered}$$

When the car is moving with just one half  $v_{\max }$, the force available is given by
$P_{\max }=F\left(\frac{v_{\max }}{2}\right)\therefore F=\frac{P_{\max }}{\left(\frac{v_{\max }}{2}\right)}=\frac{7460}{\left(\frac{3.73}{2}\right)}$

$Hencea=\frac{F-R}{m}=\frac{4000-2000}{1000}=2m/s^2$