Instantaneous (Midpoint)/Average Velocity?

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In summary, the conversation discusses proving two theorems related to average and instantaneous velocity using kinematic equations. The first theorem is proved using the equation X-Xo = Vot + 1/2at^2 and V=Vo + at, while the second theorem is attempted to be proved using X-Xo = 1/2 (V+Vo)t and V(t/2) = Vo + a(t/2). The conversation also mentions the need to relate these theorems to calculate kinetic energy.
  • #1
cudah
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Homework Statement



Prove the following theorems:
Theorem 1:

Vavg= (Vfinal+Vinitial)/2

TO Theorem 2:

Vavg=Vmp= V(t/2) <--Where V(t/2) represents a function read as V of t/2.
(t/2) is the midpoint in time for a given interval and V(t/2) is the instantaneous velocity at that time

Homework Equations



I need to prove these theorems esp theorem 2 (need to prove theorem 1 first) to relate average velocity (measured in lab) to instantaneous velocity (needed to calculate kinetic energy)

The Attempt at a Solution


I was able to prove theorem 1 using kinematic equation:
X-Xo = Vot + 1/2at^2 and V=Vo + at
Solution:
X-Xo = (2Vot)/2 + (at^2)/2
X-Xo = (2Vot + at^2)/2
(X-Xo)/t = (2Vo + at)/2
(X-Xo)/t = [(Vo + at) + Vo]/2, substitute V=Vo + at
to get,
(X-Xo)/(t-to)= (V + Vo)/2= Vavg

For theorem 2, I have to prove it using a kinematic equation too. I tried but I'm not sure if I'm doing it right.
I tried using X-Xo = 1/2 (V+Vo)t
to get,
X-Xo= [(V+Vo)t]/2
(X-Xo)/(V+Vo) = t/2

Or using V(t/2) = Vo + a(t/2) ----> V of t/2
and plug V(t/2) in X-Xo = 1/2 (V+Vo)t
X-Xo = 1/2 [{(Vo + a(t/2)} +Vo)]t
but my answer didn't make sense.
 
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  • #2
This is with constant acceleration, right?

Try drawing a graph of displacement vs time, and mark on all the variables that are mentioned in your (correct) solution to theorem 1. That should help you to think your way around the second part.
 
Last edited:
  • #3
I'm not sure if I'm doing it right.


Both Theorem 1 and Theorem 2 relate average velocity to instantaneous velocity, but in different ways. Theorem 1 simply states that the average velocity is equal to the sum of the final and initial velocities divided by 2. This can be proven using the kinematic equations as shown in your solution.

Theorem 2, on the other hand, introduces the concept of the midpoint in time, represented by t/2. This midpoint is important because it represents the exact moment at which the instantaneous velocity is being measured. To prove Theorem 2, you can use the equation V(t) = Vo + at, where V(t) is the velocity at a specific time t. You can then substitute t/2 for t to get V(t/2) = Vo + a(t/2). This equation represents the instantaneous velocity at the midpoint in time.

To further prove Theorem 2, you can use the equation X-Xo = Vot + 1/2at^2 and substitute V(t/2) for V to get X-Xo = V(t/2)t + 1/2a(t/2)^2. Simplifying this equation will give you X-Xo = 1/2(V+Vo)t. This is the same equation you used in your attempt, but you need to use the instantaneous velocity V(t/2) instead of the average velocity Vavg.

In conclusion, both Theorem 1 and Theorem 2 are valid ways to relate average velocity to instantaneous velocity. Theorem 1 is a simple equation that can be easily proven using kinematic equations, while Theorem 2 introduces the concept of the midpoint in time and uses the equation V(t) = Vo + at to prove the relationship between average and instantaneous velocity.
 

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

Instantaneous velocity refers to the velocity of an object at a specific moment in time, while average velocity is the average rate of change of an object's position over a given time interval.

2. How is instantaneous velocity calculated?

Instantaneous velocity can be calculated by finding the derivative of an object's position function with respect to time. In other words, it is the slope of the position-time graph at a specific point.

3. Can instantaneous velocity ever be negative?

Yes, instantaneous velocity can be negative if the object is moving in the negative direction (such as moving backwards). It can also be negative if the object is slowing down (decreasing in speed) in the positive direction.

4. How is average velocity different from average speed?

Average velocity takes into account the direction of motion, while average speed does not. Average speed is calculated by dividing the total distance traveled by the total time taken, while average velocity is calculated by dividing the displacement (change in position) by the total time taken.

5. What are some real-life examples of instantaneous and average velocity?

An example of instantaneous velocity could be the speed of a car at a specific moment during a trip. An example of average velocity could be the speed of a runner during a race, where their average velocity would be their total distance traveled divided by the total time taken.

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