1 A car of mass 1000 kg moves along a horizontal straight road, passing through points \(A\) and \(B\). The power of its engine is constant and equal to 15000 W . The driving force exerted by the engine is 750 N at \(A\) and 500 N at \(B\). Find the speed of the car at \(A\) and at \(B\), and hence find the increase in the car's kinetic energy as it moves from \(A\) to \(B\).

$[15000 = 750v_A, 15000 = 500v_B]$ | M1 | For using $P = Fv$
Speeds are $20\text{ms}^{-1}$ and $30\text{ms}^{-1}$ | A1 |
$[\text{KE gain} = \frac{1}{2}1000(30^2 - 20^2)]$ | M1 | For using $\text{KE} = \frac{1}{2}mv^2$
Increase is $250,000\text{J}$ (or $250\text{kJ}$) | A1ft | 4 | ft $500(v_a^2 - v_a^2)$

1 A car of mass 1000 kg moves along a horizontal straight road, passing through points $A$ and $B$. The power of its engine is constant and equal to 15000 W . The driving force exerted by the engine is 750 N at $A$ and 500 N at $B$. Find the speed of the car at $A$ and at $B$, and hence find the increase in the car's kinetic energy as it moves from $A$ to $B$.

\hfill \mbox{\textit{CAIE M1 2009 Q1 [4]}}