Question

A person is running at his maximum speed of 4m/s to catch a train. What he is 6m from the door of the compartment the train starts to leave the station at a constant acceleration of 1m/s^2. Find how it takes him to catch up the train

Answer 1

The time taken by the man to catch the train is 2 seconds

Explanation:

Let the time taken by the person to catch the train is t

In time t if the train covers d distance then the man has to cover distance = (d+6) m

Since man moves with constant speed of 4 m/s

Therefore,

$d+6=4\times t$

$\implies d=4t-6$

Again since the train starts from rest with constant acceleration of 1 m/s²

Therefore, using the second equation of motion

$d=0\times t+\frac{1}{2}\times 1\times t^2$

$\implies 4t-6=\frac{t^2}{2}$

$\implies 8t-12=t^2$

$\implies t^2-8t+12=0$

$\implies t^2-6t-2t+12=0$

$\implies t(t-6)-2(t-6)=0$

$\implies (t-2)(t-6)=0$

$\implies t=2, 6$

So here we are getting two values of t, both are valid

Since in 2 seconds the person will reach the gate

Therefore, the person will catch the train in 2 seconds

If he doesn't catch the train and keep running then due to the acceleration of the train, the speed of the train will keep increasing and then again the door of the compartment will be surpassing him after 6 seconds.

Hope this answer is helpful.

Know More:

Q: A man runs at a speed of 4m/s to overtake a standing bus when he is 6 m bend the door the bus moves forward and continues with constant acceleration of 1.2 m/s.s the man shall gain the bus at time:

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Q: A  man runs at a constant speed of 4 m/s to overtake a standing bus. When  he is 6.0 m behind the door (t=0), the bus moves forward and continues  with a constant acceleration of 1.2 m/s2.

a) How long does it take for the man to reach the door?

b) If in the beginning he is at 10.0 m from the door of the buss, will he

ever catch up?

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