# A locomotive is accelerating at 2.33m/s2. It passes through a 20.0m wide crossing in a time of 2.00s. After the locomotive leaves the crossing, how much time is required until its speed reaches...

A locomotive is accelerating at 2.33m/s2. It passes through a 20.0m wide crossing in a time of 2.00s. After the locomotive leaves the crossing, how much time is required until its speed reaches 40.0m/s?

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A locomotive is accelerating at 2.33m/s2. It passes through a 20.0-m-wide crossing in a time of 2.00s.

Let the initial velocity of the locomotive be u. So,

`s = ut + 1/2at^2`

`rArr 20 = 2u +1/2 * 2.33 * 2^2`

`rArr 20 = 2u + 4.66`

`rArr 2u=15.34`

`rArr u = 7.67` m/s

So it's at 7.67 m/s just as it finishes crossing. Now find the time until it reaches 40m/s:

`v = u + at`

`rArr 40 = 7.67 + 2.33t`

`rArr 2.33t = 32.33`

`rArr t= 13.875` s

Therefore, the required time is **13.875 s.**

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Here in the equation

`s =v_0*t +(a*t^2)/2`

`V_0` is the initial speed (not the final speed after all the space s has been travelled). Therefore in the above solution `V_0=7.67 m/s` is the speed at the beginning of the crossing.

The total time since the speed is 7.67 m/s until it is 40 m/s is of course (as above) `t=13.875 seconds` , but from this time one need to substract 2 seconds (the time that the locomotive needs to pass through the crossing).

**The correct answer is** `t =13.875-2 =11.875 seconds`

Sure, Valentin68, I accept that.

Sorry for the overlooking!

The final answer should be 2.0 seconds less, i.e. 11.875 s