How Is Average Velocity Half of Final Velocity in Constant Acceleration?

[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.

Thanks in advance

 

[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...

 

[cookiemonster]

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?

cookiemonster

 

[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

 

[NateTG]

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

[tex]\vec{x}=\vec{x}_0+\frac{\vec{v}_0+\vec{v}}{2}t[/tex]

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

 

[HallsofIvy]

Assuming constant acceleration, a, then the speed after time t is
v_f= v₀+ at (so that t= (v_f-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 v_f-v₀)/a.