Displacement equation due to acceleration

[Phys_Boi]

Using the formulas: s = [tex] \frac{1}{2} \alpha t^2 [/tex]

v = [tex] \frac{d}{t} [/tex]
a = [tex] \frac{v}{t} [/tex]​

When we divide distance "s" by time we get velocity:

v = [tex] \frac{\frac{1}{2} \alpha t^2}{t} [/tex] = [tex] \frac{1}{2} \alpha t [/tex]
​

When we divide velocity "v" by time we get acceleration:

a = [tex] \frac{\frac{1}{2} \alpha t}{t} [/tex] = [tex] \frac{1}{2} \alpha [/tex]​

½a ≠ a

​

 

[Khashishi]

You need to use calculus here. The instantaneous velocity is not the same as the average velocity.

 

[Phys_Boi]

You need to use calculus here. The instantaneous velocity is not the same as the average velocity.

So the velocity = the derivative with respect of time of acceleration?

 

[Khashishi]

You got that backwards. Acceleration is derivative of velocity wrt time

 

[Phys_Boi]

You got that backwards. Acceleration is derivative of velocity wrt time

So does that make v = ∫a ?

 

[Khashishi]

yeah, plus an integration constant

 

[Phys_Boi]

yeah, plus an integration constant

Thank you.

 

[russ_watters]

Note that under constant acceleration, the calculus is easy and you can probably even see without calculus that the average speed under a linear acceleration is half the final speed.