Question

It takes the skytrain two minutes to travel from Brentwood Station to Holdom Station. Let t be the time (in minutes) after 12:00 noon. Suppose the train leaves Brentwood Station at 11:59 am (so t=-1), and arrives at 12:01 pm (so t=1). The speed of the train at time t is exactly f(t) kilometers per hour, where
f(t)=(120/1+t^2)-60, for -1 ≤ t ≤ 1.

a) What is the maximum speed of the train when traveling between the stations?


b) Assuming that the track is straight, what is the distance between the two stations? ( Hint: The speed of the train at time t in kilometers per minutes is the function g(t)=f(t)/60).

c) What is the average speed, vave, of the train when traveling between the stations? Express your answer in kilometers per hour.

Answer 1

(a) find where df/dt = 0
or, just note that f(t) is an even function.

(b) since f(t) is the velocity, the distance is
∫[-1/60,1/60] f(t) dt


(c) avg speed is distance/time