Question

A person is running at his
maximum speed of 2 m/s to
watch a train. When he is 6m from
the door of the compartment the
train starts to leave the station at
a constant acceleration of 1m/s.
Time taken by him to catch up the
train is​

Answer 1

Explanation:

Let t be time taken.

During this time, the man travels 2t.

During time t the train travels 0×t + (1/2) (1) t^2.

Distance travelled by man = 6+ distance by train

2t = (1/2)t^2 +6

Multiply each term by 2 : t^2+12 = 4t

t^2–4t +12= 0

Here the discriminant is (-4)^2 -4•1• 12= -32 = Negative

Hence no solution. The man can't catch the train.

Answer 2

The man cannot catch the train.

Given: A person is running at his maximum speed of 2 m/s to watch a train. When he is 6 m from the door of the compartment the train starts to leave the station at a constant acceleration of 1 m/s.

To Find: The time taken by him to catch up the train.

Solution:

We can solve the numerical by using the formula of motion,

             S = ut + 1/2 × a × t²                                              .....(1)

Where S = distance, u = initial velocity, a = acceleration, t = time.

Coming to the numerical,

The initial velocity (u) is = 2 m/s

The relative acceleration = - 1 m/s²

The distance to be covered = 6 m

So, putting respective values in (1), we get;

                S = ut + 1/2 × a × t²      

           ⇒  6 = 2t + ( 1/2 ) × ( -1 ) × t²

           ⇒   t² - 4t + 12 = 0                                                    ...(2)

In equation (2), we check the discriminant,

               4² - ( 4 × 1 × 12 )

          =  16 - 48

          =   - 32 < 0

Since the discriminant of equation (2) comes out to be negative so there is no solution for the equation.

Hence, the man cannot catch the train.

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