Average and instantaneous velocity

In summary, the conversation discusses finding the average velocity on the interval [1,3] using the formula S(t) = t/(1+t^2). The first method uses the formula Vavg = (S(t0) - S(t1)) / (t0 - t1) and the second method takes the derivative of S(t) to find instantaneous velocities at t=1 and t=3, and then averages them. However, the second method is incorrect as it does not take into account the different notions of average velocity and thus produces a different answer. The correct answer is -1/10.
  • #1
physicsernaw
41
0

Homework Statement


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

Homework Equations


Vavg = (S(t0) - S(t1)) / (t0 - t1)

or

Vavg = (V0 + V1)/2

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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  • #2
physicsernaw said:

Homework Statement


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

Homework Equations


Vavg = (S(t0) - S(t1)) / (t0 - t1)

or

Vavg = (V0 + V1)/2


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.
 
  • #3
Dick said:
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?
 
  • #4
physicsernaw said:
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.
 
  • #5
Dick said:
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!
 

1. What is the difference between average velocity and instantaneous velocity?

Average velocity is defined as the total displacement of an object over a given time interval, while instantaneous velocity is the velocity of an object at a specific moment in time. In other words, average velocity gives an overall picture of an object's motion, while instantaneous velocity provides information about the object's motion at a specific point in time.

2. How is average velocity calculated?

Average velocity is calculated by dividing the total displacement of an object by the total time it took to travel that distance. So, the formula for average velocity is average velocity = (change in displacement) / (change in time).

3. What is the unit of measurement for velocity?

The unit of measurement for velocity is distance per unit time, such as meters per second (m/s) or kilometers per hour (km/h).

4. Can instantaneous velocity be negative?

Yes, instantaneous velocity can be negative. This means that the object is moving in the opposite direction of its initial motion. For example, if an object is moving to the right with an initial velocity of 10 m/s, its instantaneous velocity could be -5 m/s if it starts moving to the left at a certain moment in time.

5. Is it possible for average velocity to be the same as instantaneous velocity?

Yes, it is possible for average velocity and instantaneous velocity to be the same. This occurs when an object is moving at a constant speed in a straight line, as the instantaneous velocity at any point in time would be equal to the average velocity over a given time interval.

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