Question

Example 20. A man is s=9 m behind the door of a train
when it starts moving with acceleration a = 2 ms-2. The man
runs at full speed. How far does he have to run and after
what time does he get into the train ? What is his full speed ?​

Answer 1

Explanation:

First of all the acceleration of the train is 2 m per second “square”

Let the man catch the train after t s .

Initial velocity of the train = 0 m/s

Acceleration of the train = 2m/s

Final velocity of the train when the man catches the train = 0 + 2×t = 2t

Distance travelled by the train in t s = 1/2 ×2× t 2

= t 2

Let the velocity of the man be v m/s .

Distance travelled by the man to catch the train in t s = v×t

Initial separation between him and the train = 9 m

Since the man catches the train distance travelled by him equals to initial separation between him and the train + distance travelled by the train.

i.e. v×t = 9 + t 2 …….. eqn. i

When the man catches the train the velocity of the man must equalize that of the train and since the man runs with full velocity from the beginning of his journey, his velocity must be equal to the final velocity of the train.

Thus , v = 2t

From eqn. i

2t×t = 9 + t 2

=> 2t 2 = 9 + t 2

=> t 2 = 9

=> t = 3

Thus Full velocity of the man = 2×3 = 6 m/s

and Distance travelled by him = 9+ 3 2 = 18 m

Please mark as brainliest answer!!!

Answer 2

Answer:

example 20

Explanation:

Example 20. A man is s=9 m behind the door of a train

when it starts moving with acceleration a = 2 ms-2. The man

runs at full speed. How far does he have to run and after

what time does he get into the train ? What is his full speed .?