1. ## Kinematics (Deriving Equation)

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

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

My Problem with this question:

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

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

My Problem with this question:
the retardation force is initially $-5m N$ and is increasing uniformly with distance.

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

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

the rest follows.

Bobak

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

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

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

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

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

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

-Dan

4. ^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, $-40m$ and $20m$. I accepted $20m$ which is correct but realised that they are asking for distance so doesn't that mean that it could also be $|-40|m = 40m$?

5. 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, $-40m$ and $20m$. I accepted $20m$ which is correct but realised that they are asking for distance so doesn't that mean that it could also be $|-40|m = 40m$?
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.

6. 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 $-40m$ then $+20m$ (displacement) so the displacement is $-20m$. But as they are asking for distance, it would be the modulus of $|-20m|$ hence the answer being $20m$?

7. Originally Posted by Air
Could it have been that it travelled $-40m$ then $+20m$ (displacement) so the displacement is $-20m$. But as they are asking for distance, it would be the modulus of $|-20m|$ hence the answer being $20m$?
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 $v^2 = 400 - 10x - \frac{x^2}{2}$.

1. v = 20 m/s. Then $v = + \sqrt{400 - 10x - \frac{x^2}{2}}$, 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 $v = - \sqrt{400 - 10x - \frac{x^2}{2}}$, 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 $v^2 = 400 - 10x - \frac{x^2}{2}$ is an ellipse.