Question

A police car traveling at 95.0 km/h is traveling west, chasing a motorist traveling at 80.0 km/h.

(a) What is the velocity of the motorist relative to the police car?

(b) What is the velocity of the police car relative to the motorist?

(c) If they are originally 250 m apart, in what time interval will the police car overtake the motorist?

Short answer

(a) The velocity of the motorist relative to police car is 15.0 km/h towards east.

(b) The velocity of the police car relative to the motorist is 15.0 km/h towards west.

(c) The time interval required by the police to overtake the motorist is$0.0167 \text{h}=60.0 \text{s}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of the police car is, 95.0 km/h in west.
• The speed of the motorist is, 80.0 km/h .

The velocity of the motorist relative to the police car 

The velocity of the motorist relative to the police car is,

$$\begin{gathered} {\mathit{v}}_{\mathbf{m}\mathbf{p}}\mathbf{=}{\mathit{v}}_{\mathbf{m}\mathbf{w}}\mathbf{+}{\overrightarrow{\mathbf{v}}}_{\mathbf{w}\mathbf{p}} \\ {\mathit{v}}_{\mathbf{m}\mathbf{p}}\mathbf{=}{\mathit{v}}_{\mathbf{m}\mathbf{w}}\mathbf{-}{\overrightarrow{\mathbf{v}}}_{\mathbf{p}\mathbf{w}} \end{gathered}$$

Here,

${\mathit{v}}_{\mathbf{mp}}$is the velocity of the motorist relative to the police car.

${\overrightarrow{\mathbf{v}}}_{\mathbf{pw}}$ is the velocity of the police car relative to west direction.

${\mathit{v}}_{\mathbf{mw}}$is the velocity of the motorist relative to west direction.

Determination of the velocity of the motorist relative to the police car

(a)

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{v}_{\text{mp}}& =& 80.0 \text{km/h}+(-95.0 \text{km/h})\\ & =& -15.0 \text{km/h}\\ & & \end{array}$

The negative sign shows that the relative velocity of the motorist is towards east.

Conclusion:

Therefore, the velocity of the motorist relative to police car is $15.0 \text{km/h}$towards east.

The velocity of the police car relative to the motorist

The velocity of the police car relative to the motorist is,

$v_{pm}=-v_{mp}$

Here, $v_{pm}$is the velocity of the police car relative to the motorist.

From part (a) the velocity of the motorist relative to the police car is,

${\text{v}}_{\text{mp}}=-15.0 \text{km/h}$

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{v}_{\text{pm}}& =& -(-15.0 \text{km/h})\\ & =& 15.0 \text{km/h}\end{array}$

Therefore, the velocity of the police car relative to the motorist is 15 km/h towards west.

The time interval required by the police to overtake the motorist

(c)

From part (b) the velocity of the police car relative to the motorist is 15 km/h towards west.

The time taken by the police car to overtake the motorist is,

$t=\frac{d}{v_{pm}}$

Here,

d is the distance between the police car and the motorist.

t is the time taken by the police car to overtake the motorist.

Substitute the value in the above equation,

$\begin{array}{rcl}t& =& \frac{250\times {10}^{-3} \text{km}}{15.0 \text{km/h}}\\ & =& 0.0167 \text{h}\\ & =& 0.0167 \text{h}\left(\frac{3600 \text{s}}{1 \text{h}}\right)\\ t& =& 60.0 \text{s}\end{array}$

Therefore, the time interval required by the police to overtake the motorist is$0.0167 \text{h}=60.0 \text{s}$.