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.
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
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?
Yes, you are on the right track. To find the distance between the two stations, you need to integrate the speed function, f(t), from t = -1 to t = 1. However, there seems to be an error in the calculation you provided.
Let's evaluate the integral step by step:
First, we need to calculate the speed of the train as a function of minutes, g(t). We know that g(t) = f(t)/60. Therefore:
g(t) = (f(t))/60
= ((120/(1+t^2)) - 60)/60
= (2/(1+t^2)) - 1
Now, let's integrate g(t) from t = -1 to t = 1:
∫[(-1 to 1)] g(t) dt = ∫[(-1 to 1)] ((2/(1+t^2)) - 1) dt
To evaluate this integral, we can use a standard integral formula or method, like substitution. Applying the formula, we get:
∫[(-1 to 1)] ((2/(1+t^2)) - 1) dt = [2arctan(t) - t] ∣[(-1 to 1)]
Evaluating the definite integral, we get:
[2arctan(t) - t] ∣[(-1 to 1)] = [2arctan(1) - 1] - [2arctan(-1) - (-1)]
Since arctan(1) = π/4 and arctan(-1) = -π/4, we can simplify further:
[2arctan(1) - 1] - [2arctan(-1) - (-1)] = [2(π/4) - 1] - [2(-π/4) + 1]
= (π/2) - 1 + (π/2) + 1
= π
Therefore, the distance between the two stations is π kilometers.