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?
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 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.
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