Question

a cyclist is moving on road at 1 m per second parallel to railway track a train coming from behind when start crossing him he start his stop clock he finds that train crosses him in 10 second if length of train is 200 m find speed of train assuming uniform

Answer 1

HELLO THERE!

This numerical can be solved using the Relative Velocity Concept.

Note, that since the length of the train is 200 metres, the train has to travel 200 metres to cross the cycle, at the relative velocity.

Let the train be moving with a velocity of x m/s. Relative velocity of Train with respect to Cyclist = x - 1 (because the cycle is moving at 1 m/s).

So, the train covers 200 metres in 10 seconds at a speed of (x-1) m/s.

Using equation of motion:

$S = ut + \frac{1}{2}at^{2}$

We know that a = 0, because the body is moving with uniform velocity (as given in the question).

So,

[tex] S = ut \\\\\implies 200 = (x-1)\times10 \\\\\implies x-1=20 \\\\\implies x = 21 [/tex]

Hence, speed of the train is 21 m/s.

HOPE MY ANSWER IS SATISFACTORY...

Correct me if I'm wrong :)

Answer 2

$Answer:$

The question given below needs to be solved by the concept of Relative Velocity.

Given, length of train to be 200 metres.

So in order to cross the cycle, the train has to travel 200 metres.

Now, let the train move with a velocity of x m/s.

So the relative velocity of train with respect to cyclist = (x-1) [cyclist is moving with an acceleration of 1 m/s.

Therefore, the distance travelled by train in 10 sec will be 200 metres and that too with a velocity of (x-1) m/s.

Acceleration will be zero, since it's given that the train is in uniformely accelerated motion.

$Using \: second \: equation \: of \: motion , \\ \\ s = ut + \frac{1}{2} {at}^{2}$

$= > 200 = (x - 1) \times 10 \\ \\ = > x - 1 = 20 \\ \\ = > x = 21 \: \frac{m}{s}$

Hence speed will be 21 m/s.