A high speed passenger test train travels from Columbia station to Penn station in exactly 8 hours. The distance traveled in miles from Columbia, at any given time in hours is given by the following function: s(t)=6t^3+72t^2

What is the average speed of the train (average rate of change) between 2 and 5 hours?

What is the velocity (instantaneous rate of change) of the train at exactly 2.5 hours into the trip?

What is the maximum velocity attained by the train? Please justify.

2. Originally Posted by Hallah
A high speed passenger test train travels from Columbia station to Penn station in exactly 8 hours. The distance traveled in miles from Columbia, at any given time in hours is given by the following function: s(t)=6t^3+72t^2

What is the average speed of the train (average rate of change) between 2 and 5 hours?

What is the velocity (instantaneous rate of change) of the train at exactly 2.5 hours into the trip?

What is the maximum velocity attained by the train? Please justify.
Hello,

a) The average speed is: $\displaystyle \frac{s(5)-s(2)}{5-2}= 738 \ \frac{m}{h}$

b) The velocity is: $\displaystyle v(t)=\frac{ds}{dt}=18t^2+144t$
So v(2.5) = 472.5 m/h

c) You get the maximum velocity if $\displaystyle \frac{dv}{dt}=36t+144=0$
So you get t = -4.

As I've commented on your other post (concerning the same ghost train) the given function cann't be correct, because this train accelerates all the time. In my opinion a train should reach a constant speed pretty soon and hold it during the whole trip.

Greetings

EB