# How to prove v(t) = (y(t+Δt) - y(t-Δt))/2Δt

• sabastronomia
In summary, to prove that v(t) = (y(t+Δt) - y(t-Δt))/2Δt using kinematic equations for constant acceleration, one can use the equation y(t) = y(0) + v(0)t + at^2/2 and evaluate y(t+Δt) and y(t-Δt), subtract them, and divide by 2Δt. Another approach is to use y(t) = y(0) + v(t)t - at^2/2 and manipulate it to solve for v(t). Both methods use the concept of average velocity and the definition of acceleration to prove the given equation.
sabastronomia
1. Prove that:
v(t) = (y(t+Δt) - y(t-Δt))/2Δt

using kinematic equations for constant acceleration.

3. I tried using the limit as Δt approaches 0 of (y(t+Δt)-y(t))/Δt and somehow adding to get Δt. Could I manipulate the kinematic equations somehow by setting t equal to (t+Δt) or (t-Δt)?

What's the normal definition of v(t) in terms of these sorts of limits?

sabastronomia said:
1. Prove that:
v(t) = (y(t+Δt) - y(t-Δt))/2Δt

using kinematic equations for constant acceleration.

3. I tried using the limit as Δt approaches 0 of (y(t+Δt)-y(t))/Δt and somehow adding to get Δt. Could I manipulate the kinematic equations somehow by setting t equal to (t+Δt) or (t-Δt)?

Since acceleration is constant: y(t+Δt) = y(t) + vavgΔt = y(t) + (v(t) + v(t+Δt))Δt/2

Does that help?

AM

Try using the equation:

y(t)=y(0)+v(0)t+at2/2

Evaluate y(t+Δt) and y(t-Δt), subtract them, and divide by 2Δt, and see what you get.

Chestermiller said:
Try using the equation:

y(t)=y(0)+v(0)t+at2/2

Evaluate y(t+Δt) and y(t-Δt), subtract them, and divide by 2Δt, and see what you get.

Perhaps:

y(t)=y(0)+v(t)t-at2/2

might help as well.

## 1. What is v(t) and how is it related to y(t)?

V(t) represents the velocity of an object at a specific time, t. It is related to y(t) by the equation v(t) = (y(t+Δt) - y(t-Δt))/2Δt, where y(t+Δt) and y(t-Δt) represent the positions of the object at times t+Δt and t-Δt, respectively.

## 2. How is the equation v(t) = (y(t+Δt) - y(t-Δt))/2Δt derived?

This equation is derived from the definition of velocity, which is the change in position over the change in time. By using the positions at two different times, t+Δt and t-Δt, and dividing by the change in time, 2Δt, we can calculate the average velocity at time t.

## 3. What does Δt represent in the equation?

Δt represents the change in time, or the time interval between t+Δt and t-Δt. It is usually a small value, as we want to calculate the average velocity at a specific time t.

## 4. Can this equation be used for non-uniform motion?

Yes, this equation can be used for non-uniform motion as long as we have the positions of the object at two different times, t+Δt and t-Δt. However, it will only give us the average velocity at time t, not the instantaneous velocity.

## 5. How can we prove that this equation is accurate?

To prove the accuracy of this equation, we can use it to calculate the average velocity at different time intervals and compare it to the actual velocity of the object at that time. If the calculated average velocity closely matches the actual velocity, then we can conclude that the equation is accurate.

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