Kinematics (Deriving Equation)

**Simplicity** · May 30th 2008, 02:44 AM

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:
I can't derive it.

Help please? Thanks in advance.

**bobak** · May 30th 2008, 02:59 AM

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:
I can't derive it.

Help please? Thanks in advance.

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

**topsquark** · May 30th 2008, 03:06 AM

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:
I can't derive it.

Help please? Thanks in advance.

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

**Simplicity** · May 30th 2008, 07:03 AM

^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 $|-40|m = 40m$ ?

**mr fantastic** · May 30th 2008, 07:59 PM

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.

**Simplicity** · May 31st 2008, 02:03 AM

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 $20m$ ?

**mr fantastic** · May 31st 2008, 03:29 AM

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.

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