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.
`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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