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Question

A commuter train travels between two downtown stations. Because the stations are only 1.12 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 = -0.600 m/s2 for a time interval Δt2. Find the minimum time interval of travel Δt and the time interval Δt1.

Answer 1

I assume the train starts from rest at the first station, and must return to rest (come to a complete stop) at the second station.

Although this sounds like a calculus problem that is asking you to find the value of Δt1 that minimizes the total travel time, it really isn't.  The problem is completely constrained by the values given in the question, and it's really just an algebra problem.

The maximum speed reached by the train is:

v_max = a1 * Δt1

and because it must come to rest at the second station, we also know that:

0 = v_max + a2 * Δt2 = a1 * Δt1 + a2 * Δt2

so

Δt2 = -(a1/a2)*Δt1

Now the total distance covered is D (= 1.12 km)

D = 0.5*a1*(Δt1)^2 + (v_max*Δt2) + 0.5*a2*(Δt2)^2

The first term in the above equation is the distance covered while the train is accelerating at a1, and the next two terms are the distance covered while the train is decellerating at a2.

Plugging in the expressions for V_max and Δt2 as functions of Δt1, we get:

D = 0.5*a1*(Δt1)^2 + a1*Δt1*(-(a1/a2)*Δt1) + 0.5*a2*(-(a1/a2)*Δt1)^2

D = 0.5*a1*(Δt1)^2 - ((a1^2)/a2)*Δt1^2 + 0.5*((a1^2)/a2)*(Δt1)^2

D = 0.5*a1*(Δt1)^2 - 0.5*((a1^2)/a2)*Δt1^2

D = 0.5*(a1 - (a1^2)/a2)*Δt1^2

D = 0.5*((a1*a2 - (a1^2))/a2)*Δt1^2

Δt1^2 = (2*D*a2)/(a1*a2 - a1^2)

Plugging in the appropriate numbers gives:

Δt1^2 = 19200 sec^2

Δt1 = 138.564 sec

Using the equation, Δt2 = -(a1/a2)*Δt1 we find that

Δt2 = (.1/.6)*(138.564 s*x) = 23.094 sec

So the total time elapsed is

Δt = Δt1 + Δt2 = 138.564 + 23.094 = 161.658 sec