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In summary, the formula d=vt+1/2at^2 can be derived by calculating the average velocity and using the equations for displacement and acceleration. It can also be derived using calculus by integrating the equations for velocity and acceleration.
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anandzoom
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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?
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berkeman
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anandzoom said:
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.
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anandzoom
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berkeman said:
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)
- #4
berkeman
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anandzoom said:
See AlsoWhy is the distance calculation multiplied by one half in this physics problem?Understanding the D = Vit + 1/2at^2 Formula | Tips and Tricksv=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?
- #5
anandzoom
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berkeman said:
Thank you, that helps. Have you had any Calculus yet? Are you familiar with derivatives?
Yup
- #6
berkeman
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anandzoom said:
Yup
Great. In that case, where did the bolded equation come from below?
anandzoom said:
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)
- #7
anandzoom
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berkeman said:
Great. In that case, where did the bolded equation come from below?
acceleration= velocity/time
- #8
berkeman
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anandzoom said:
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:
- #9
anandzoom
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How can you say that displacement is area under the velocity curve?
- #10
berkeman
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anandzoom said:
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"...
[tex]x = \int{v(t) dt}[/tex]
- #11
anandzoom
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Ok thanks
- #12
rcgldr
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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
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- #13
guedman
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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
Related to How do I correctly derive the formula d = vt + 1/2 at^2?
1. How do you derive the equation D=vt+1/2at^2?
To derive the equation D=vt+1/2at^2, we start with the definition of acceleration, which is the rate of change of velocity over time. This can be represented as a=Δv/Δt. We can rearrange this to solve for velocity, giving us v=aΔt. We then substitute this value for velocity into the equation for distance, D=vt, giving us D=aΔtΔt. Simplifying this, we get D=at^2. Since this equation only accounts for constant acceleration, we add in the initial velocity term, giving us the final equation D=vt+1/2at^2.
2. What does each term in the equation D=vt+1/2at^2 represent?
The D in the equation represents the total distance traveled, v is the initial velocity, t is the time, a is the constant acceleration, and 1/2at^2 represents the change in distance due to acceleration. Essentially, this equation combines the equations for constant velocity (D=vt) and constant acceleration (D=1/2at^2) into one formula.
3. How is the equation D=vt+1/2at^2 used in physics?
The equation D=vt+1/2at^2 is commonly used in physics to determine the distance an object has traveled under constant acceleration. This is useful in a variety of scenarios, such as calculating the distance a car travels during braking or the distance a ball travels when thrown into the air.
4. Can the equation D=vt+1/2at^2 be used for non-constant acceleration?
No, the equation D=vt+1/2at^2 is only valid for constant acceleration. If the acceleration is changing, the equation must be modified to account for this. However, in cases where the acceleration is changing at a constant rate, this equation can still be used by breaking the motion into smaller intervals with constant acceleration.
5. Are there any limitations to using the equation D=vt+1/2at^2?
Yes, there are several limitations to using this equation. Firstly, it only applies to motion with constant acceleration. Additionally, it assumes that there is no external force acting on the object, and that the acceleration is in the same direction as the initial velocity. It also does not take into account other factors such as air resistance or friction, which can affect the motion of an object. Finally, this equation is only applicable in the absence of relativistic effects.
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