Average and instantaneous velocity

physicsernaw

1. Homework Statement
Given the following formula for distance, find the average velocity on the interval [1,3]
S(t) = t/(1+t^2)

2. Homework Equations
Vavg = (S(t0) - S(t1)) / (t0 - t1)

or

Vavg = (V0 + V1)/2

3. The Attempt at a Solution

I get two different answers and I need help understand why.

Vavg = (S(1) - S(3)) / (1-3)

Vavg = (1/2 - 3/10) / -2

Vavg = -2/20 = -1/10

Now using a different method, getting the instantaneous velocity at t = 1 and t = 3 by taking derivative of S(t) -> S'(t) = (1-t^2)/(1+t^2)^2

Vavg = (S'(1) + S'(3))/2

Vavg = (-8/100)/2 = (-2/25)/2 = -1/25

Why am I getting different averages here?? I don't understand what I'm doing wrong.

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Dick

Homework Helper
1. Homework Statement
Given the following formula for distance, find the average velocity on the interval [1,3]
S(t) = t/(1+t)^2

2. Homework Equations
Vavg = (S(t0) - S(t1)) / (t0 - t1)

or

Vavg = (V0 + V1)/2

3. The Attempt at a Solution

I get two different answers and I need help understand why.

Vavg = (S(1) - S(3)) / (1-3)

Vavg = (1/2 - 3/10) / -2

Vavg = -2/20 = -1/10

Now using a different method, getting the instantaneous velocity at t = 1 and t = 3 by taking derivative of S(t) -> S'(t) = (1-t^2)/(1+t^2)^2

Vavg = (S'(1) + S'(3))/2

Vavg = (-8/100)/2 = (-2/25)/2 = -1/25

Why am I getting different averages here?? I don't understand what I'm doing wrong.
Judging by your work, I'm guessing you meant S(t)=t/(1+t^2), not S(t)=t/(1+t)^2. And you can't average instantaneous velocities to get average velocities. The first one is right, the second one isn't.

physicsernaw

Judging by your work, I'm guessing you meant S(t)=t/(1+t^2), not S(t)=t/(1+t)^2. And you can't average instantaneous velocities to get average velocities. The first one is right, the second one isn't.
Yes I'm sorry that's what I meant. But I don't understand, why can't you do what I did? If (v0 + v) / 2 = average velocity, and v0 and v are velocities at an instant in time, I don't understand how the second method doesn't work?

Dick

Homework Helper
Yes I'm sorry that's what I meant. But I don't understand, why can't you do what I did? If (v0 + v) / 2 = average velocity, and v0 and v are velocities at an instant in time, I don't understand how the second method doesn't work?
Because they are two totally different notions of average velocity. If I start at rest in city A and drive to city B and park, then my beginning and ending velocities are zero. My average velocity, in the sense you want, isn't zero.

physicsernaw

Because they are two totally different notions of average velocity. If I start at rest in city A and drive to city B and park, then my beginning and ending velocities are zero. My average velocity, in the sense you want, isn't zero.
Got it, thank you!

"Average and instantaneous velocity"

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