Question

The driver of train A moving at a constant speed of $30m/s$ sees another train B moving on the track at a speed of $10m/s$ in the same direction. He immediately applies break and achieves a uniform retardation of 3m/s². To avoid collision, what must be the minimum distance between the trains?

Answer 1

Step 1: Given data

The final velocity of train A, $v=0m/s$

Initial velocity of train A, $u=30m/s$

Acceleration of train A $a=-3m{s}^2$

Velocity of train B, $v_b=10m/s$

Let s, t, and X are the distance covered by train A, time, and distance covered by train B respectively.

Step 2: Formula used

$v^2=u^2+2as$ (third equation of motion)

$v=u+at$ (First equation of motion)

$Distance=speed\times time$

Step 3: Finding the time to stop

Putting values

$0=30-3t$

$\Rightarrow t=10seconds$

Step 4: Finding the distance covered before stopping

$v^2=u^2+2as$

$0=30\times 30-2\left(3\right)s$

$s=900/6=150m$

Step 5: Finding the time train B would have covered

$X=10\times 10=100m$

Step 6: Finding the minimum distance between the trains

Therefore minimum distance = $150-100=50m$.
Hence, the minimum distance between the trains is $50m$.