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

[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)?

 

[PeroK]

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

 

[Andrew Mason]

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) + v_avgΔt = y(t) + (v(t) + v(t+Δt))Δt/2

Does that help?

AM

 

[Chestermiller]

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.

 

[PeroK]

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.