# D=vt+1/2at^2 derivation

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1. Apr 1, 2015

### anandzoom

d=vt+1/2at^2
Derive me this formula please... When I do I get 1/2(vt+at^2)...i know that I'm doing a mistake by not considering the initial velocity... So how do I correct it?

Last edited by a moderator: Apr 1, 2015
2. Apr 1, 2015

### Staff: Mentor

Welcome to the PF.

Can you show us the steps of your derivation? That will help us find any errors.

3. Apr 1, 2015

### anandzoom

v=d/t ;d=vt
a=d/t^2 ;d=at^2
1/2d+1/2d=1/2vt+1/2at^2
d=1/2(vt+at^2)

4. Apr 1, 2015

### Staff: Mentor

Thank you, that helps. Have you had any Calculus yet? Are you familiar with derivatives?

5. Apr 1, 2015

Yup

6. Apr 1, 2015

### Staff: Mentor

Great. In that case, where did the bolded equation come from below?

7. Apr 1, 2015

### anandzoom

acceleration= velocity/time

8. Apr 1, 2015

### Staff: Mentor

Actually it's the change in velocity over time, or in calculus, a(t) = dv(t)/dt.

But it's also possible to do the derivation for a constant acceleration using only triangles. Here's a YouTube video that helps to explain it:

9. Apr 1, 2015

### anandzoom

How can you say that displacement is area under the velocity curve?

10. Apr 1, 2015

### Staff: Mentor

That's from calculus:

v(t) = dx(t)/dt

a(t) = dv(t)/dt

So when you integrate both sides of the first equation, you are effectively finding the "area under the curve"...

$$x = \int{v(t) dt}$$

11. Apr 1, 2015

### anandzoom

Ok thanks

12. Apr 1, 2015

### rcgldr

If acceleration is constant, then you can use algebra instead of calculus. Note that Δt means change in time, Δx means change in position, a = acceleration, v = velocity.

initial velocity = v
final velocity = v + a Δt
average velocity = (initial velocity + final velocity) / 2 = ((v) + (v a Δt))/2 = v + 1/2 a Δt
Δx = average velocity Δt = (v + 1/2 a Δt) Δt = v Δt + 1/2 a Δt^2

Last edited: Apr 1, 2015
13. Apr 2, 2015

### guedman

it is not possible to write v = d/t but v=dx/dt and a = dv/dt
then v(t) =∫a(t)dt =at + v , v is the initial velocity and the acceleration a is constant.
it comes : x(t) =∫v(t)dt = 1/2 at^2 + vt + x(0), x(0) is the initial position assumed nill then x(0) =0

finally the distance d is x(t) =d= vt + 1/2 a t^2