Question

78.  A commuter train travels between two downtown stations. Because the stations are only 1.00 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the time interval ?t₁ between two stations by accelerating the rate a₁ = 0.100 m/s² for a time interval ?t₁ and then immediately braking with acceleration a₂ = ?0.500 m/s² for a time interval ?t₂. Find the minimum time interval ?tand the time interval ?t_1.

Answer 1

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

The maximum speed reached by the train is:

v_max = a1*t1

and because it must come to rest at the second station,

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

so,
t2 = -(a1/a2)*t1

Now the total distance covered is D (= 1 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 decelerating 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 = 16000

t1 = 126.5sec


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

t2 = (.1/.4)*(126.494 sec) = 31.62 sec

So the total time elapsed is

t = t1 + t2 = 126.5+31.62 = 158.12.