Find velocity-time equation from velocity-position equation

[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 $v=a(t-t_{0})+v_{0}$ ...

 

[haruspex]

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 $v\frac{dv}{dx}=\frac{dv}{dt}$.

 

[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 $v\frac{dv}{dx}=\frac{dv}{dt}$.

Thank you for your time.
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 $v(t=0)$.

 

[PeroK]

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 $v =u+at$.

 

[PeroK]

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.

 

[PeroK]

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?

 

[PeroK]

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 $(t_2,v(t_2))$?

 

[PeroK]

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.

Thank you for your time.