91Ó°ÊÓ

In this problem, the frame of reference in which the train is steady is considered. First, for calculating the time to pass the train, you will have to find the relative velocity of the car to the train and then use the constant horizontal velocity relation to determine the time required.

Given data:

The velocity of the car is $v_1=95\mathrm{km}/\mathrm{h}$.

The length of the train is $d=1\mathrm{km}$.

The velocity of the train is role="math" localid="1644923596041" $v_2=75\mathrm{km}/\mathrm{h}$.

When the car and train are moving in the same direction, the formula to calculate the time can be written as:

$v_x=\frac{d}{t_1}$

Here, $v_x$is the velocity of the car relative to the train, and $t_1$is the time taken by the car.

On plugging the values in the above relation, you get:

$\begin{array}{c}{v}_1-v_2=\left(\frac{1\mathrm{km}}{t_1}\right)\\ 95\mathrm{km}/\mathrm{h}-75\mathrm{km}/\mathrm{h}=\left(\frac{1\mathrm{km}}{t_1}\right)\\ t_1=\left(0.05\mathrm{h}×\frac{60×60\mathrm{s}}{1\mathrm{h}}\right)\\ t_1=180\mathrm{s}\end{array}$

Thus,$t_1=180\mathrm{s}$ is the time taken by the car to pass the train.

The formula to calculate the distance traveled can be written as:

$D=v_1{t}_1$

On plugging the values in the above relation, you get:

$\begin{array}{c}D=\left(95\mathrm{km}/\mathrm{h}\right)\left(0.05\mathrm{h}\right)\\ D=4.75\mathrm{km}\end{array}$

Thus, $D=4.75\mathrm{km}$is the distance covered by the car to pass the train.

When the car and train are moving in opposite directions, the formula to calculate the time can be written as:

$v_x\text{'}=\frac{d}{t_2}$

Here, $v_x\text{'}$is the velocity of the car relative to the train, and $t_2$is the time taken by the car.

On plugging the values in the above relation, you get:

$\begin{array}{c}{v}_1+v_2=\left(\frac{1\mathrm{km}}{t_2}\right)\\ 95\mathrm{km}/\mathrm{h}+75\mathrm{km}/\mathrm{h}=\left(\frac{1\mathrm{km}}{t_2}\right)\\ t_2=\left(\frac{1}{170}\mathrm{h}×\frac{60×60\mathrm{s}}{1\mathrm{h}}\right)\\ t_2=21.17\mathrm{s}\end{array}$

Thus, $t_2=21.17\mathrm{s}$is the time taken by the car to pass the train.

The formula to calculate the distance traveled can be written as:

$D\text{'}=v_1{t}_2$

On plugging the values in the above relation, you get:

$\begin{array}{c}D\text{'}=\left(95\mathrm{km}/\mathrm{h}\right)\left(\frac{1}{170}\mathrm{h}\right)\\ D\text{'}=0.55\mathrm{km}\end{array}$

Thus, $D\text{'}=0.55\mathrm{km}$is the distance covered by the car to pass the train.