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Find the acceleration of the train and show that the total distance travelled is 302.5 metres, when the rear of the train is at B.

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Consider the following quantities:

*a *is the acceleration of the train

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 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 + 50t`

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`

Then `V_1 = 11 - 10a = 11 - 10* 1/5 = 9 m/s`

Now from the very first equation we could find the time it took the head of the train to reach signal post B:

`t=V_1/a = 11/(1/5) = 55 s`

That means the time it took the rear of the train to reach signal post B is 10 seconds greater: 55 s + 10 s = 65 s

That means the total distance traveled by the time the rear of the train reaches B is

`d = at^2/2 = (1/5)*(65)^2 / 2 = 422.5 m`

This is the distance traveled by the head of the train. The rear of the train traveled 100 m less:

422.5 m - 100 m = 322.5 meters

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