# Why are avg. v and instant. v different?

If x(b)= 20-15b+2b^2 then why would average and instantaneous velocity values be the same?

If x(b)= 20-15b+2b^3 then why would average and instantaneous velocity values be different? What would the magnitude of the difference mean?

I think in x(b)= 20-15b+2b^2 average and instantaneous velocity values are the same because the formula for average velocity calculates the slope of two b centered at the b between the two different b. While the formula for instantaneous velocity is the line for the slope of t, which is also the center of two different t.
But then it wouldn't make sense for x(b)= 20-15b+2b^3. Basically I don't have any idea for the second equation.

If x(b)= 20-15b+2b^2 then why would average and instantaneous velocity values be the same?

If x(b)= 20-15b+2b^3 then why would average and instantaneous velocity values be different? What would the magnitude of the difference mean?

I think in x(b)= 20-15b+2b^2 average and instantaneous velocity values are the same because the formula for average velocity calculates the slope of two b centered at the b between the two different b. While the formula for instantaneous velocity is the line for the slope of t, which is also the center of two different t.
But then it wouldn't make sense for x(b)= 20-15b+2b^3. Basically I don't have any idea for the second equation.

For instantaneous velocity we take the limit as Δt approaches to zero.
For average velocity we can take Δt for any duration.

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So i could not understand what that b represents....

But simply, if u have displacement as a function of time, u can use

average speed = total displacement/total time

and instantaneous velocity = df(t)/dt
that is differentiate ur (displacement as a function of time) function to get instantaneous velocity

The speed (obviously) changes at certain times, due to acceleration. The average velocity takes the average speed at all points divided by the time (makes sense right?). The instantaneous velocity takes the exact point, therefore it's the derivative of the displacement.

Thank you for your help and everyone makes sense but what I don't understand is why for some equations the average velocity and instantaneous velocity are equal and some aren't. In this case the average velocity for both equations is calculated for t=4 seconds centered at t=2. But for the first equation the average and instantaneous velocity are the same for t=2,4,6, etc... while for the second the equation the average and instantaneous velocity are different with a magnitude of 8. I can't explain why.

Chestermiller
Mentor
In my judgement, you had it doped out correctly in the first place. You are talking about using a finite difference approximation to estimate the instantaneous velocity. For a parabolic shape to a curve, the central difference formula gives the exact velocity at the center of the time interval. For a cubic variation along the curve, the central difference formula is not exact, and there is a discretization error involved. For the specific cubic variation under consideration, evaluate the velocity at b and at b + $\Delta$b, and then calculate the average velocity by dividing by $\Delta$b. Then use the derivative to evaluate the velocity at the center of the interval b + ($\Delta$b)/2. Then subtract the two results to determine the discretization error. If you do this procedure for the parabolic variation, you will always find that the discretization error is zero.

rcgldr
Homework Helper
This is similar to mean value theorem. For any continous function x(t) over the interval {a, b}, there is at least one point c within the interval {a,b} where x'(c) = (x(b) - x(a)) / (b - a). This means there is some point c where the instantaneous velocity x'(c) equals the average velocity.

For the second equation

x(t) = 2 t3 - 15 t + 20

the average velocity over {0,t} is

(x(t) - x(0)) / (t - 0) = 2 t2 - 15

define a so that (a t) = c, the value we're looking for

the instantaneous velocity is

x'(a t) = 6 (a t)2 - 15

then find a so that instantaneous velocity = average velocity
x'(a t) = (x(t) - x(0)) / (t - 0)

6 (a t)2 - 15 = 2 t2 - 15
6 (a)2 t2 = 2 t2
6 (a)2 = 2
(a)2 = 1/3
a = sqrt(1/3)

so c = sqrt(1/3) t

This means that for x(t) = 2 t3 - 15 t + 20, the instantaneous velocity at sqrt(1/3) t equals the average velocity over {0,t}.

You've already noted that for x(t) = 2 t3 - 15 t + 20, the instantaneous velocity at (1/2) t equals the average velocity over {0,t}.

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