Question

Commuter train travels between two downtown stations. because the stations are only 1.18 km apart, the train never reaches its maximum possible cruising speed. during rush hour the engineer minimizes the travel interval δt between the two stations by accelerating for a time interval δt1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.350 m/s2 for a time interval δt2. find the minimum time interval of travel δt and the time interval δt1.

Answer 1

Distance between stations = 1180 m

We have v = u+at

When the train is accelerating v = 0+0.100*δt1 = 0.1δt1

When the train is decelerating 0 = 0.1δt1 - 0.350*δt2

So δt1 = 3.5 δt2

Now we have s = ut +
`(1)/(2) at^2`

For the accelerating part

S1 = 0*δt1 +
`(1)/(2)*0.1* (\delta t1)^2 = 0.05* (\delta t1)^2`

For the decelerating part

S2 =
`0.1* \delta t1*\delta t2-(1)/(2)* 0.35*(\delta t2)^2`

We have S1+S2 = 1180

`1180 = 0.05*(\delta t1)^2+0.1(\delta t1)(\delta t2) - (1)/(2)*0.35*(\delta t2)^2\\ \\ 1180 = 0.6125*(\delta t2)^2+0.35* (\delta t2)^2-0.175*(\delta t2)^2\\ \\ 0.7875*(\delta t2)^2 = 1180\\ \\ \delta t2 = 38.71 seconds\\ \\ \delta t1 = 135.485 seconds\\ \\ \delta t = 174.195seconds`