An unmarked police car traveling a constant$\mathbf{95}\mathbf{}{\mathbf{kmh}}^{\mathbf{-}\mathbf{1}}$is passed by a speeder traveling$\mathbf{135}\mathbf{}{\mathbf{kmh}}^{\mathbf{-}\mathbf{1}}$. Precisely 1.00s after the speeder passes, the police officer steps on the accelerator; if the police car’s acceleration is$\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$, how much time passes before the police car overtakes the speeder (assumed to be moving at constant speed)?

The motion of an object is said to be uniformly accelerated when its velocity changes at a constant rate with respect to time.

Initial velocity of the police car,$v_{\mathrm{p}}=95{\mathrm{kmh}}^{-1}$

Velocity of the speeder,$v_{\mathrm{s}}=135{\mathrm{kmh}}^{-1}$

Acceleration of police car, $a=2.60{\mathrm{ms}}^{-2}$

Let the time taken by the police car to catch the speeder be t.

The velocity of the police car in meters per second is.

$\begin{array}{c}{v}_{\mathrm{p}}=95\times \frac{1000}{3600}\\ =26.4{\mathrm{ms}}^{-1}\end{array}$

Similarly, the velocity of the speeder in meters per second is.

$\begin{array}{c}{v}_{\mathrm{s}}=135\times \frac{1000}{3600}\\ =37.5{\mathrm{ms}}^{-1}\end{array}$

As the speeder moves with a constant velocity, the distance he covers in the given time is.

$\begin{array}{c}{d}_{\mathrm{s}}=v_{\mathrm{s}}t\\ =37.5t\end{array}$

The police car moves at a constant speed for the initial 1 s, and after that, it accelerates at $2.60{\mathrm{ms}}^{-2}$.

From the second equation of motion, the distance it covers in time t is.

$d_{\mathrm{p}}=\left(v_{\mathrm{p}}\times 1\right)+\left(v_{\mathrm{p}}\times \left(t-1\right)+\frac{1}{2}\times a\times {\left(t-1\right)}^2\right)$

Substituting the values in the above equation,

role="math" localid="1642850312994" $d_{\mathrm{p}}=\left(26.4\times 1\right)+\left(26.4\times \left(t-1\right)+\frac{1}{2}\times 2.60\times {\left(t-1\right)}^2\right)$

The above equation can be simplified as.

$d_{\mathrm{p}}=1.3t^2+23.8t+1.3$

As the police catch the speeder at the end, they both travel the same distance.

$d_{\mathrm{s}}=d_{\mathrm{p}}$

Substituting the values from previous results,

$37.5t=1.3t^2+23.8t+1.3$

The above equation can be re-written as.

$1.3t^2-13.7t+1.3=0$

Solving the quadratic equation,

$t=10.44\mathrm{s}$

(Ignoring the other root, which was approximately equal to zero)

Thus, the police car takes 10.44 s to catch the speeder.