Question

A speeder passes a parked police car with a speed of 65 km/h in a 50km/h zone. One second after the speeder has passed the police car, the police begin his pursuit. The police car accelerates with constant acceleration of 2m/s^2.Required:a. How long does it take for the police to catch the speeding car? b. How far did the police car travel before police caught up with the speeder? c. What is the speed of the police car when catches up with the speeder?

Answer 1

Answer:

a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.

Explanation:

Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.

a) How long does it take for the police to catch the speeding car:

Kinematic equation of each vehicle's position are described:

Speeder

$s_{A} = s_{A,o}+v_{A}\cdot t$

Police Car

$s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}$

If $s_{A} = s_{B}$, $s_{A,o} = s_{B,o}$, $v_{A} = 18.056\,\frac{m}{s}$, $v_{B,o} = 0\,\frac{m}{s}$ and $a_{B} = 2\,\frac{m}{s^{2}}$, the resulting expression is done:

$v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}$

$\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0$

$t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0$

$t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}$

$t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }$

$t = 18.056\,s$

The police will take 18.056 seconds to catch the speedy car.

b) How far did the police car travel before police caught up with the speeder?

The distance travelled by the police is: ($s_{B,o} = 0\,m$,  $a_{B} = 2\,\frac{m}{s^{2}}$, $v_{B,o} = 0\,\frac{m}{s}$ and  $t = 18.056\,s$)

$s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}$

$s_{B} = 326.019\,m$

The police will travel 326.019 meters before catching the speedy car.

c) What is the speed of the police car when catches up with the speeder?

The speed of the police car is represented by the following formula:

$v_{B} = v_{B,o} + a_{B}\cdot t$

Where $v_{B}$ is the speed of the police car, measured in meters per second.

Given that $v_{B,o} = 0\,\frac{m}{s}$, $t = 18.056\,s$ and $a_{B} = 2\,\frac{m}{s^{2}}$, the final speed of the police car when catches up with the speeder is:

$v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)$

$v_{B} = 36.112\,\frac{m}{s}$

The speed of the police car when catches up with the speeder is 36.112 meters per second.