Thread: Kinematics (Deriving Equation)

Question:
A particle starts with speed $\displaystyle 20 \ ms^{-1}$ and moves in a straight line. The particle is subject to a retardation which is initially $\displaystyle 5 \ ms^{-2}$ and which increases uniformly with the distance moved, having a value of $\displaystyle 11 \ ms^{-2}$ when the particle has moved $\displaystyle 12 \ m$. Given that the particle has speed $\displaystyle v \ ms^{-1}$ and has moved a distance of $\displaystyle x$, show that, while the particle is in motion

$\displaystyle v \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right)$.

My Problem with this question:
I can't derive it. Help please? Thanks in advance.

Originally Posted by Air

Question:
A particle starts with speed $\displaystyle 20 \ ms^{-1}$ and moves in a straight line. The particle is subject to a retardation which is initially $\displaystyle 5 \ ms^{-2}$ and which increases uniformly with the distance moved, having a value of $\displaystyle 11 \ ms^{-2}$ when the particle has moved $\displaystyle 12 \ m$. Given that the particle has speed $\displaystyle v \ ms^{-1}$ and has moved a distance of $\displaystyle x$, show that, while the particle is in motion

$\displaystyle v \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right)$.

My Problem with this question:
I can't derive it. Help please? Thanks in advance.

the retardation force is initially $\displaystyle -5m N$ and is increasing uniformly with distance.

so the retardation force is of the form $\displaystyle -(5+kx)m$. Your given when $\displaystyle x = 12$ and the force has magnitude $\displaystyle -11m$

Finding k is pretty simple. then you have $\displaystyle m \ddot{x} = -(5+kx)m $

the rest follows.

Bobak

Originally Posted by Air

Question:
A particle starts with speed $\displaystyle 20 \ ms^{-1}$ and moves in a straight line. The particle is subject to a retardation which is initially $\displaystyle 5 \ ms^{-2}$ and which increases uniformly with the distance moved, having a value of $\displaystyle 11 \ ms^{-2}$ when the particle has moved $\displaystyle 12 \ m$. Given that the particle has speed $\displaystyle v \ ms^{-1}$ and has moved a distance of $\displaystyle x$, show that, while the particle is in motion

$\displaystyle v \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right)$.

My Problem with this question:
I can't derive it. Help please? Thanks in advance.

We know the speed decreases uniformly with the distance, so
$\displaystyle \frac{da}{dx} = \text{constant} = \frac{-11~m/s^2 - -5~m/s^2}{12~m} = -\frac{1}{2}~s^{-2}$

Thus
$\displaystyle a = a_0 + \frac{da}{dx} x$

$\displaystyle a = -5 - \frac{1}{2}x = - \left ( 5 + \frac{1}{2}x \right )$

Now for a little variable work:
$\displaystyle a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$
so we are done.

-Dan

^Thanks.

Also, a follow on question is:
Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

I worked it out to get two answers, $\displaystyle -40m$ and $\displaystyle 20m$. I accepted $\displaystyle 20m$ which is correct but realised that they are asking for distance so doesn't that mean that it could also be $\displaystyle |-40|m = 40m$?

Originally Posted by Air

^Thanks.

Also, a follow on question is:
Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

I worked it out to get two answers, $\displaystyle -40m$ and $\displaystyle 20m$. I accepted $\displaystyle 20m$ which is correct but realised that they are asking for distance so doesn't that mean that it could also be $\displaystyle |-40|m = 40m$?

So you found the value of x when v = 0?

While the particle continues to move in the same direction (slowing down of course) the value of x when v = 0 will be the distance travelled (assuming the initial condition x = 0 at t = 0). So x = -40 is a spurious solution.

Originally Posted by mr fantastic

So you found the value of x when v = 0?

While the particle continues to move in the same direction (slowing down of course) the value of x when v = 0 will be the distance travelled (assuming the initial condition x = 0 at t = 0). So x = -40 is a spurious solution.

Could it have been that it travelled $\displaystyle -40m$ then $\displaystyle +20m$ (displacement) so the displacement is $\displaystyle -20m$. But as they are asking for distance, it would be the modulus of $\displaystyle |-20m|$ hence the answer being $\displaystyle 20m$?

Originally Posted by Air

Could it have been that it travelled $\displaystyle -40m$ then $\displaystyle +20m$ (displacement) so the displacement is $\displaystyle -20m$. But as they are asking for distance, it would be the modulus of $\displaystyle |-20m|$ hence the answer being $\displaystyle 20m$?

No.

I'll take moving right as the positive direction and moving left as the negative direction. There are two cases to consider: v = 20 m/s (the particle moves to the right with an intial speed of 20 m/s) and v = -20 m/s (the particle moves to the left with an initial speed of 20 m/s). You have $\displaystyle v^2 = 400 - 10x - \frac{x^2}{2}$.

1. v = 20 m/s. Then $\displaystyle v = + \sqrt{400 - 10x - \frac{x^2}{2}}$, x > 0.

Draw a graph of v versus x. Note that it starts at v = 20 and x = 0 (corresponding to t = 0) and x > 0. It's very clear that when v = 0, x = 20. That is, the distance travelled is 20 m.

2. v = -20 m/s. Then $\displaystyle v = - \sqrt{400 - 10x - \frac{x^2}{2}}$, x < 0.

Draw a graph of v versus x. Note that it starts at v = -20 and x = 0 (corresponding to t = 0) and x < 0. It's very clear that when v = 0, x = -40. That is, the distance travelled is 40 m.


The origin of your two solutions should now be clear. Clearly the book has assumed v = 20 m/s .......


By the way, the graph of $\displaystyle v^2 = 400 - 10x - \frac{x^2}{2}$ is an ellipse.