Question

It takes the Skytrain two minutes to travel from Brentwood Station to Holdom Station. Let t br the time (in minutes) after 12:00 noon. Suppose the train leaves Brentwood Station at 11:59 am (t= -1), and arrives at 12:01 pm (t= 1). The speed at time t is exactly f(t) kilometres per hour, where
f(t)= (120/1+t^2) - 60, for -1 ≤ t ≤ 1
Assuming that the track is straigh, what is the distance between the two stations? (hint; The speed of the train at time t in kilometres per minute is the function g(t) = f(t) /60.)

Am I supposed to take the integral of f(t) from -1 to 1? I ended up getting a negative number which was -3600km

Thanks in advance.

Answer 1

The next part of the question is:
What is the average speed of the train when traveling between the stations? Express your answer in kilometres per hour.
What formula should I use for this question?

Answer 2

yes, distance is the integral of velocity. So,
v(t) = 120/(1+t^2) - 60
in km/min, that is 2/(1+t^2) - 1
Thus, the distance traveled is
s(t) = ∫[-1,1] v(t) = 2 arctan(t) - t [-1,1]
= (2(π/4)-1)-(2(-π/4)+1) = 1/π km/min
Too bad you didn't show your integral.
avg speed is distance/time, so that would be (1/π)/2 = 1/2π km/min = 30/π km/hr