Question

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)?

Short answer

The police car overtakes the speeder after 10.4 s from the time the speeder passes by the police car.

Step-by-step solution

Step 1. Definition of uniformly accelerated motion

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

Step 2. Data identification and assumption

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.

Step 3. Conversion of the unit of velocity

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.

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

Step 4. Calculation of the distance covered by the speeder in time t

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

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

Step 5. Calculation of the distance covered by the police car in time t

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.

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

Step 6. Calculation of the value of t

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.