Question

The driver of a train moving with a constant speed $v_1$ along a straight track sights another train at a distance d ahead of him on the same track moving in the same direction with a constant speed $v_2$. He at once applies the brakes and gives his train a constant retardation f. There will be a collision of the trains if:

• A. $v_1>v_2$ and $\frac{(v_1-v_2{)}^2}{2f}<d$
• B. $v_1<v_2$ and $\frac{(v_1+v_2{)}^2}{2f}>d$
• C. $v_1>v_2$ and $\frac{(v_1-v_2{)}^2}{2f}>d$
• D. $v_1>v_2$ and $\frac{(v_1^2-v_2^2)}{2f}>d$

Answer 1

The correct option is C $v_1>v_2$ and $\frac{(v_1-v_2{)}^2}{2f}>d$

Relative velocity $v_{12}=v_1-v_2$,

Relative separation $s_{12}=d$,

Relative acceleration $a_{12}=a_1-a_2=-f$

Let at time $t$ , trains collides.

Using $s=ut+\frac{1}{2}a{t}^2$ $\Rightarrow d=(v_1-v_2)t-\frac{1}{2}f{t}^2$

$\Rightarrow f{t}^2-2(v_1-v_2)t+2d=0$,

For real solution of $t$, $\left(2\right(v_1-v_2){)}^2-4\times f\times 2d>0$

$\Rightarrow \frac{(v_1-v_2{)}^2}{2f}>d$

$\therefore$ For collision $v_1>v_2$ and $\frac{(v_1-v_2{)}^2}{2f}>d$