Question

A train is moving at a constant speed V when its driver observes another train in front of him on the same track and moving in the same direction with constant speed v. If the distance between the trains is x, then what should be the minimum retardation of the train so as to avoid collision?

• A. $\frac{(V+v{)}^2}{x}$
• B. $\frac{(V-v{)}^2}{x}$
• C. $\frac{(V+v{)}^2}{2x}$
• D. $\frac{(V-v{)}^2}{2x}$

Answer 1

The correct option is D $\frac{(V-v{)}^2}{2x}$

Train 1 velocity $=v$

Train 2 velocity $=u$

Let distance between trains be $x$

Relative velocity of $t_2$ with repect to $t_1$ will be,

$u_{t_2{t}_1}=v_{t_2}-v_{t_1}$

$=u-v$

According to Newtons 3rd law of motion

$v^2-u^2=2as$

$a=\frac{v^2-u^2}{2s}$

since final velocity is 0 then,

$s=\frac{0-(u-v{)}^2}{2x}$

$a=\frac{(v-u{)}^2}{2x}$