- #1
crazyecto
- 1
- 0
The graph below shows a plot of velocity vs. time for an object undergoing uniformly accelerated motion. The object has instantaneous velocity v1 at time t1 and instantaneous velocity v2 at time t2. Use the graph and the fact that, when the acceleration is constant, the average velocity can be written as vavg = (v1 + v2)/2 to explain how we know that the instantaneous velocity at the midpoint time is equal to vavg on the interval.
V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t
Now I know the answer has to do with some how deriving one of these formulas to get it to equal the formula for average veolcity but I can't seem to remember how. I think you some how take the derivative of X - Xo = .5(Vo + V)t to get the answer maybe?
V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t
Now I know the answer has to do with some how deriving one of these formulas to get it to equal the formula for average veolcity but I can't seem to remember how. I think you some how take the derivative of X - Xo = .5(Vo + V)t to get the answer maybe?