# Average and Final Velocity

1. Mar 9, 2004

### xxpsychoxx

I got this problem and it just stumped me. Can anyone give me the solution? Anyways, here's the question:
Show that the average velocity of a body undergoing constant acceleration, and starting from rest. is half of its final velocity.

2. Mar 9, 2004

### Physicsisfun2005

i was trying to solve your problem with some made up #'s but i couldn't get the V average to equal Vf. It may be my methods.....

3. Mar 9, 2004

Sounds like you're already given the solution. The trick is getting there, right?

Why don't you show us what you've tried so far?

4. Mar 9, 2004

### blackflub

well if you take the initial V plus the final V and average it out your going to get the average V. (0+Vf)/2=1/2Vf I hope that helps

5. Mar 9, 2004

### NateTG

There's a nifty equation for constant acceleration which essentially gives you the answer.

$$\vec{x}=\vec{x}_0+\frac{\vec{v}_0+\vec{v}}{2}t$$

You can also derive it from:
$$\vec{x}=\vec{x}_0+\vec{v}_0t+\frac{1}{2}\vec{a}t^2$$
and
$$\vec{v}=\vec{v}_0+\vec{a}t$$
by solving the bottom equation for $$t$$ and substituting in to the top one.

6. Mar 10, 2004

### HallsofIvy

Staff Emeritus
Assuming constant acceleration, a, then the speed after time t is
vf= v0+ at (so that t= (vf-v0)/a ) and the distance moved is v0t+ (1/2)at2.

At constant speed, u, the distance moved would be
ut. The average speed must move you the same distance as the actual speed in time t: ut= v0tf+ (1/2)at. Solve for u, then replace t by vf-v0)/a.