Average Value//Velocity/Acceleration Question

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Hi-- im stuck on this problem

A car starts from rest (=0 and t=0) and experiences constant acceleration x''(t)=a for T seconds. Find, in terms of a and T, (a) its final and average velocities and (b) its final and average positions..

so V avg is integral from 0 to t Vdx/t which is just position over time... and position average is integral from 0 to t of position/time.. idk im really confused, i asked my prof a question about it and said that we need exact answers? but the problem says in terms of a and T, so who knows.. im confused. Any suggestions? Thanks :)
 
Slowly, Carefully, One Step at a Time.

s"(t) = v'(t) = a(t) = a
s'(t) = v(t) = a*t + V0

"starts from rest " - So V0 = 0

s'(t) = v(t) = a*t
s(t) = ½*a*t^2 + H0

"rest (=0 and t=0)" - So H0 = 0

s(t) = ½*a*t^2

Time T

Final Velocity: v(T) = a*T
Final Position: s(t) = ½*a*T^2

What about those averages?
 
isnt V avg just position/time? and Position average = velocity* time? but it has to be in terms of a and T.... so im not sure how to get it in that form
 
aswimmer113 said:
isnt V avg just position/time? and Position average = velocity* time? but it has to be in terms of a and T.... so im not sure how to get it in that form
Click to expand...
No, but you are very close.

Average Velocity is (CHANGE in Position) / (Change in Time)

Ask yourself, "Where did it start," and "Where did it stop?"
 
The average of f on [a,b] is \(\displaystyle Ave_{f,[a,b]} = \frac{{\int_a^b {f(x)dx} }}{{b - a}}.\)

Now, you just apply that simple idea to these problems.
Your interval is [0,T] one function is v(t)=at the other is s(t)=(1/2)at<SUP>2</SUP>.
 
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