How do I correctly derive the formula d = vt + 1/2 at^2?

[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?

 

[berkeman]

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?

Welcome to the PF.

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

 

[anandzoom]

Welcome to the PF.

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

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)

 

[berkeman]

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)

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

 

[anandzoom]

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

Yup

 

[berkeman]

Yup

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

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)

 

[anandzoom]

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

acceleration= velocity/time

 

[berkeman]

acceleration= velocity/time

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:

 

[anandzoom]

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

 

[berkeman]

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

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"...

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

 

[anandzoom]

Ok thanks

 

[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

 

[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