Find velocity-time equation from velocity-position equation

• MatinSAR
In summary: 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.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
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}##.

Last edited:
PeroK and MatinSAR
haruspex said:
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)##.

MatinSAR said:
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
PeroK said:
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##.

MatinSAR said:
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
PeroK said:
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?

MatinSAR said:
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
PeroK said:
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))##?

MatinSAR said:
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
PeroK said:
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.

PeroK

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