Instantaneous velocity from avg velocity with constant accelartion

[KeilB]

Homework Statement

With constant acceleration prove that the average velocity from t₁ to t₂ =t₁ + Δt is equal to the instanous velocity in the middle of the time interval between t1 and t2.

Homework Equations

What I am looking for is a general equation that does not involve accelaration. All I have is position. By the way this is 1-D kinematics so no need for vectors.

The Attempt at a Solution

Tried all sorts of subs with Vavg= (Vi + Vf)/2 and Vavg= Δx/Δt I tried subbing these into many of the other equations such as Vf=Vi+at. I imagine there is some sort of trick to it. This thing is driving me nuts.

 

[Ninty64]

Vavg= (Vi + Vf)/2
Vf=Vi+at

It can be shown this way I believe. Here's a hint:
Vavg= (Vi + Vf)/2
Vavg= (Vi + Vf)/2 - Vi +Vi
Vavg= (Vf - Vi)/2 +Vi

 

[KeilB]

Alright still having a hard time. I got it into that form that you mentioned and was able to get Vavg=2Vf-(3at)/2 which there is a t/2 in there. Good sign I suppose but it just doesn't seem right since I am going to have to use a distance traveled to find it. I will play around with it some more but feel I'm coming to dead ends.

 

[vela]

Can you express in math what it is exactly that you're trying to show? In other word, what equation says "the average velocity from t₁ to t₂ is equal to the instantaneous velocity in the middle of the time interval between t₁ and t₂"?

 

[asteriatic]

I can provide a response to this content.

First, let's define some variables:
- t1 = initial time
- t2 = final time
- Δt = change in time (t2-t1)
- Vavg = average velocity from t1 to t2
- Vi = initial velocity
- Vf = final velocity

We know that average velocity is calculated by taking the total displacement (Δx) divided by the total time (Δt):
Vavg = Δx/Δt

We also know that average velocity is equal to the instantaneous velocity at the midpoint of the time interval:
Vavg = (Vi + Vf)/2 = Vmid

Now, let's use the equation for displacement with constant acceleration:
Δx = Vit + 1/2at^2

We can substitute this into the equation for average velocity:
Vavg = (Δx)/Δt = (Vit + 1/2at^2)/Δt

Since we are looking for the instantaneous velocity at the midpoint, we can set t = (t1 + t2)/2:
Vmid = Vit + 1/2a((t1 + t2)/2)^2/Δt

Now, let's simplify this equation:
Vmid = Vi + 1/2a(t1^2 + 2t1t2 + t2^2)/2Δt
Vmid = Vi + 1/2a(t1^2 + t1t2 + t2^2)/Δt
Vmid = Vi + 1/2a(t1 + t2)(t1 + t2)/Δt
Vmid = Vi + 1/2a(t1 + t2)(Δt)
Vmid = Vi + 1/2a(Δt)^2

Now, let's go back to the equation for average velocity and substitute in our new equation for Vmid:
Vavg = (Vit + 1/2a(Δt)^2)/Δt

We can see that this is equal to Vmid, which is the instantaneous velocity at the midpoint of the time interval between t1 and t2. Therefore, we have proven that the average velocity from t1 to t2 is equal to the instantaneous velocity at the midpoint of the time interval.