91Ó°ÊÓ

The speed of an object is the scalar quantity that is defined as the rate of change of distance covered by the object in unit time.

Given data.

The speed of the trains is $v= 155 \frac{\mathrm{km}}{\mathrm{h}}$.

The distance between the trains is 8.5 km.

Assumption.

Assume that after time t, both the trains reach each other.

Assume that the first train travels $\mathrm{Δ}{x}_1$, and the second train travels $\mathrm{Δ}{x}_2$distance before reach each other.

Now, the speed of the train is

$\begin{array}{c}v =155 \frac{\mathrm{km}}{\mathrm{h}} × \frac{1000 \mathrm{m}}{1 \mathrm{km}} × \frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ = 43.06 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Now, the distance between the trains is

$\begin{array}{c}\mathrm{Δ}x = 8.5 \mathrm{km} × \frac{1000 \mathrm{m}}{1 \mathrm{km}}\\ = 8500 \mathrm{m}\end{array}$

For the first train, $\mathrm{Δ}{x}_1=vt$.

For the second train, $\mathrm{Δ}{x}_2=vt$.

Now, the total distance is

$\begin{array}{c}\mathrm{Δ}{x}_{}=\mathrm{Δ}{x}_1+\mathrm{Δ}{x}_2\\ \mathrm{Δ}{x}_{}= vt + vt\\ \mathrm{Δ}{x}_{}= 2vt\end{array}$

Now, substituting the values of $\mathrm{Δ}x$and $v$in the above equation,

$\begin{array}{c}8500\mathrm{m}= 2 × 43.06 \frac{\mathrm{m}}{\mathrm{s}} × t\\ t = 98.70\mathrm{s}\end{array}$

Therefore, they will take 98.70 s before reaching each other.