A commuter train travels between two downtown stations. Because the stations are

Question

A commuter train travels between two downtown stations. Because the stations are only 1.30 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval t between the two stations by accelerating for a time interval t1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.480 m/s2 for a time interval t2. Find the minimum time interval of travel t and the time interval t1.
Hint: It is much easier to do this problem graphically. First draw an acceleration versus time graph. What does the area under this graph physically mean? You should be able to get a relation between t1 and t2. Then draw a velocity versus time graph. What does the area under this graph physically mean? You should be able to get another relation between the two time intervals. Now you have two equations and two unknowns.

Draw position, velocity, and acceleration graphs. These must be turned in on paper.

Explanation / Answer

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 t1 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 * t1

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

0 = v_max + a2 * t2 = a1 * t1 + a2 * t2

so

t2 = -(a1/a2)*t1

Now the total distance covered is D (= 1.3 km)

D = 0.5*a1*(t1)^2 + (v_max*t2) + 0.5*a2*(t2)^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 t2 as functions of t1, we get:


D = 0.5*a1*(t1)^2 + a1*t1*(-(a1/a2)*t1) + 0.5*a2*(-(a1/a2)*t1)^2

D = 0.5*a1*(t1)^2 - ((a1^2)/a2)*t1^2 + 0.5*((a1^2)/a2)*(t1)^2

D = 0.5*a1*(t1)^2 - 0.5*((a1^2)/a2)*t1^2

D = 0.5*(a1 - (a1^2)/a2)*t1^2

D = 0.5*((a1*a2 - (a1^2))/a2)*t1^2

t1^2 = (2*D*a2)/(a1*a2 - a1^2)

Plugging in the appropriate numbers gives:


t1^2 = 21517 sec^2

t1 = 146.69 sec


Using the equation, t2 = -(a1/a2)*t1 we find that

t2 = (.1/.48)*(146.69 sec) = 30.56 sec

So the total time elapsed is

t = t1 + t2 = 146.69 + 30.56 = 177.25 sec

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