# Displacement equation due to acceleration

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1. Nov 10, 2015

### Phys_Boi

Using the formulas: s = $$\frac{1}{2} \alpha t^2$$
v = $$\frac{d}{t}$$
a = $$\frac{v}{t}$$​

When we divide distance "s" by time we get velocity:
v = $$\frac{\frac{1}{2} \alpha t^2}{t}$$ = $$\frac{1}{2} \alpha t$$
When we divide velocity "v" by time we get acceleration:
a = $$\frac{\frac{1}{2} \alpha t}{t}$$ = $$\frac{1}{2} \alpha$$​

½a ≠ a

2. Nov 10, 2015

### Khashishi

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

3. Nov 10, 2015

### Phys_Boi

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

4. Nov 10, 2015

### Khashishi

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

5. Nov 10, 2015

### Phys_Boi

So does that make v = ∫a ?

6. Nov 10, 2015

### Khashishi

yeah, plus an integration constant

7. Nov 10, 2015

### Phys_Boi

Thank you.

8. Nov 10, 2015

### Staff: Mentor

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.