# Find velocity-time equation from velocity-position equation

• MatinSAR

#### MatinSAR

Homework Statement
Find velocity-time and position-time equation from velocity-position equation(Kinematics).
Relevant Equations
##v=at+v_{0}##
##v^2=v_{0}^2+2a(x-x_{0})##
##x=x_{0}+v_{0}t+\frac 1 2 at^2##
The velocity-position equation of an object that is moving in a straight line is ##v^2-16=4x##. If ##v_{0}<0## find average velocity for ##3<t<5##.

Answer of the book starts with :

We know that :
##v^2-v_{0}^2=2a(x-x_{0})##
##v^2-16=4x##

So ##v_{0}^2=16## => ##v_{0}=-4## and ##a=2##.
##v=at+v_{0}## => ##v=2t-4##

I can't undestand how it finds out that at t=0 we have ##v_{0}=-4## because it might be ##t=t_{0}>0##. In this case book's answer is wrong be cause we should use ##v=a(t-t_{0})+v_{0}## ...

Your relevant equations are all for the constant acceleration case. You ought not assume that here. However, it does look like the book answer assumes that. It also assumes ##x_0=0##.
As to your question, the problem statement does not mention ##t_0##. It does not define ##v_0##, but it is reasonable to suppose that means the velocity at t=0.
The solver chose ##t_0## to represent t at the start, so it equals 0 by definition.

Fwiw, you can deduce acceleration is constant. Differentiating ##v^2=4x+16## wrt x yields ##2v\frac{dv}{dx}=4##, and in general ##v\frac{dv}{dx}=\frac{dv}{dt}##.

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• PeroK and MatinSAR
Your relevant equations are all for the constant acceleration case. You ought not assume that here. However, it does look like the book answer assumes that. It also assumes ##x_0=0##.
As to your question, the problem statement does not mention ##t_0##. It does not define ##v_0##, but it is reasonable to suppose that means the velocity at t=0.
The solver chose ##t_0## to represent t at the start, so it equals 0 by definition.

Fwiw, you can deduce acceleration is constant. Differentiating ##v^2=4x+16## wrt x yields ##2v\frac{dv}{dx}=4##, and in general ##v\frac{dv}{dx}=\frac{dv}{dt}##.
How can we know that ##v_0^2=16##?
It might be velocity in another time ... because in general we have ##v_{2}^2=v_{1}^2+2a(x_{2}-x_{1})## and this ##v_{1}## is different from ##v(t=0)##.

The velocity-position equation of an object that is moving in a straight line is ##v^2-16=4x##.
You can also see directly that this equation is of the form ##v^2 - (\pm 4)^2 = 2(2)x##. Compare this with the equation of constant acceleration, starting at ##x = 0##: ##v^2 - u^2 = 2ax##.

One solution, therefore, is ##a = 2, v_0 = \pm 4, x_0 = 0##.

And, let's not worry about units.

• MatinSAR
You can also see directly that this equation is of the form ##v^2 - (\pm 4)^2 = 2(2)x##. Compare this with the equation of constant acceleration, starting at ##x = 0##: ##v^2 - u^2 = 2ax##.

One solution, therefore, is ##a = 2, v_0 = \pm 4, x_0 = 0##.

And, let's not worry about units.
Thank you.
But we can use these if we have ##t_0=0##. Am I right?
I mean if for ##v_0 = \pm 4, x_0 = 0## we have ##t_0>0## the we cannot use ##v =u+at##.

we have ##t_0>0## the we cannot use ##v =u+at##.
That's purely notation. I would take ##t_0 = 0, t_1 = 3, t_2 = 5##. If you take ##t_0 = 3##, then you have the same solution, but you need to mess about unnecessarily with the standard equations.

• MatinSAR
PS if you draw a graph (extended to ##t = 0##), then it's clear. ##t_0## is just an arbitrary label on the graph. It's the same graph however you label the points, so you might as well have ##t_0 = 0##.

• MatinSAR
That's purely notation. I would take ##t_0 = 0, t_1 = 3, t_2 = 5##. If you take ##t_0 = 3##, then you have the same solution, but you need to mess about unnecessarily with the standard equations.
This is what I meant. If I choose for example ##t_0=3## then I can't use ##v=u+at##. Am I right?

This is what I meant. If I choose for example ##t_0=3## then I can't use ##v=u+at##. Am I right?
The full equations have ##x_0## and ##v_0## in them. For example:
$$v(t) = u(t_0) + a(t-t_0)$$And$$v(t_2) = u(t_1) + a(t_2 - t_1)$$and$$v(t_2)^2 - v(t_1)^2 = 2a(x(t_2) - x(t_1))$$The point of my first post here was that you should recognise an equation you've seen before. That was a quick way to realise we do have a constant acceleration solution.

• MatinSAR
PS if you draw a graph (extended to ##t = 0##), then it's clear. ##t_0## is just an arbitrary label on the graph. It's the same graph however you label the points, so you might as well have ##t_0 = 0##.
So according to this, Can I say that ##v(t)=v_0+at## is an equation of a line and we can find it by any given two points like ##(t_1,v(t_1))## and ##(t_2,v(t_2))##?

So according to this, Can I say that ##v(t)=v_0+at## is an equation of a line and we can find it by any given two points like ##(t_1,v(t_1))## and ##(t_2,v(t_2))##?
Yes, there's an important link generally between kinematic equations and geometry. The trajectory of a particle is, after all, a curve of some description in 3D space.

In a physical scenario, the starting time ##t = 0## can be literally where the motion starts (e.g. a ball being kicked). Or, it can be some arbitrary point in the trajectory (e.g. a planetary orbit).

Constant acceleration motion maps to straight lines and parabolas. Planetary and related motion maps to conic sections (circles, ellipses and hyperbolas).

A charged particle in a constant magnetic field moves in a helix.

And, more generally motion is some smooth curve in space.

• MatinSAR
Yes, there's an important link generally between kinematic equations and geometry. The trajectory of a particle is, after all, a curve of some description in 3D space.

In a physical scenario, the starting time ##t = 0## can be literally where the motion starts (e.g. a ball being kicked). Or, it can be some arbitrary point in the trajectory (e.g. a planetary orbit).

Constant acceleration motion maps to straight lines and parabolas. Planetary and related motion maps to conic sections (circles, ellipses and hyperbolas).

A charged particle in a constant magnetic field moves in a helix.

And, more generally motion is some smooth curve in space.
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