# A train, 100 metres long, starts from rest at a station A and moves with constant acceleration.Later, it takes 10 seconds to passs a singlal post B. The train is moving with velocity 11ms^-1 when...

A train, 100 metres long, starts from rest at a station A and moves with constant acceleration.Later, it takes 10 seconds to passs a singlal post B. The train is moving with velocity 11ms^-1 when the rear of the train passes B.Draw a velocitytime graph for the motion of the train.

### 2 Answers | Add Yours

The time velocity graph is attached here. If the time when front side of train meets post is t = T and velocity v = V then the rear will pass the post at t = T+10 and v = 11m/s. At this 10 seconds the train has travel 100m which is its length.

The graph will show a gradual increase since the train is accelerating.

All these data are included in the graph.

**Sources:**

Consider the following quantities:

*a *is the acceleration of the train

`V_1` is the velocity of the train when its head passes the signal post B.

Then,

`V_1 = at` , where *t* is the time it took the head of the train to reach signal post B.

Since the velocity of the train increased from `V_1` to 11 m/s in 10 seconds during which the train passed the signal post B,

`11 = V_1 + a*10` .

Since the length of the train is 100 m,

`100 = V_1*10 + (a*10^2)/2`

These three equations can be used to find *a* and *t*.

`11 = V_1 + 10a`

`100 = 10V_1 + 50a`

The second equation is equivalent to `10=V_1 + 5a`

From the first equation, `V_1=11 - 10a`

Plugging this into the second equation, get

`10=11 - 10a + 5a = 11 - 5a`

From here `a = 1/5 m/s^2`