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 1.24 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval delta t between the two stations by accelerating for a time interval delta t_1 at a_1 = 0.100 m/s^2 and then immediately braking with acceleration a_2 = -0.340 m/s^2 for a time interval delta t_2- Find the minimum time interval of travel delta t and the time interval delta t_1. 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 delta t_1 and delta t_2. 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

We 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 has been asked to find the value of t1 that minimizes the total travel time,

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 (= 1.24 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 = (2*1.24*(-0.34)/ (0.1*(-0.34) -0.1^2)) = 19163.63 sec^2

t1 = 138.43 sec

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

t2 = (.1/.34)*(138.43 sec) =40.71sec

So the total time elapsed is

t = t1 + t2 = 138.43 + 40.71 = 179.14 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.