An automobile traveling at 95 km/h overtakes a 1.30 km long train traveling in the same direction on a track parallel to the road. If the train’s speed is 75 km/h, how long does it take for the car to pass it, and how far will the car have traveled in this time? See Fig. 2-36. What are the results if the car and the train are traveling in opposite directions?

The velocity of a body is defined as the rate with which the body changes its displacement in unit time. The velocity of a body is a vector quantity, and its unit is the same as that of speed.

Given data.

The speed of the automobile (car) is

$\begin{array}{c}{\mathrm{v}}_{\mathrm{c}}= 95 \frac{\mathrm{km}}{\mathrm{h}}\\ = 95 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ = 26.39 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

The speed of the train is

$\begin{array}{c}{\mathrm{v}}_{\mathrm{t}}= 75 \frac{\mathrm{km}}{\mathrm{h}}\\ = 75 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ = 20.83 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

The automobile travels 1.30 km to pass the train.

Now, the distance is

$\begin{array}{c}1.30 \mathrm{km} = 1.30 \mathrm{km} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}}\\ = 1300 \mathrm{m}\end{array}$

Assumptions.

Let the automobile take $t$time to pass the train.

Then, the distance traveled by the automobile in time $t$is $d_{\mathrm{c}}=v_{\mathrm{c}}t$, and the distance traveled by the truck in time $t$is $d_{\mathrm{t}}=v_{\mathrm{t}}t$.

Now, after $t$time, they are in the same position. Then,

$\begin{array}{c}{v}_{\mathrm{c}}t=v_{\mathrm{t}}t+1300\mathrm{m}\\ \left(v_{\mathrm{c}}-v_{\mathrm{t}}\right)t=1300\mathrm{m}\\ \left(26.39\frac{\mathrm{m}}{\mathrm{s}}-20.83\frac{\mathrm{m}}{\mathrm{s}}\right)t=1300\mathrm{m}\\ t=233.8\mathrm{s}\end{array}$

Now, changing the unit of the time to minutes,

$\begin{array}{c}t = 233.8 \mathrm{s} \times \frac{1 \min }{60 \mathrm{s}}\\ = 3.90 \min \end{array}$

Therefore, the car takes 3.90 min to pass the train.

The traveled distance by the car within 3.90 min or 233.8 s is

$\begin{array}{c}d= 26.39 \frac{\mathrm{m}}{\mathrm{s}} \times 233.8 \mathrm{s}\\ = 6170.0 \mathrm{m}\\ = 6170.0 \mathrm{m} \times \frac{1 \mathrm{km}}{1000 \mathrm{m}}\\ = 6.17 \mathrm{km}\end{array}$

Therefore, the car has traveled 6.17 km at this time.

If they are traveling in the opposite directions,

$\begin{array}{c} v_{\mathrm{c}} t + v_{\mathrm{t}} t = 1300 \mathrm{m}\\ \left( v_{\mathrm{c}}+ v_{\mathrm{t}}\right) t = 1300 \mathrm{m}\\ \left( 26.39 \frac{\mathrm{m}}{\mathrm{s}}+ 20.83 \frac{\mathrm{m}}{\mathrm{s}}\right) t = 1300 \mathrm{m}\\ t = 27.5 \mathrm{s}\end{array}$

Therefore, when they are traveling in the opposite directions, the car takes 27.5 s to pass the train.

Now, the traveled distance by the car within 27.5 s is

$\begin{array}{c}d\text{'}= 26.39 \frac{\mathrm{m}}{\mathrm{s}} \times 27.5 \mathrm{s}\\ = 725.7 \mathrm{m}\end{array}$

Therefore, the car has traveled 725.7 m in the time when they are traveling in opposite directions.