Motion of a Particle?

Question

The acceleration of a particle is defined by the relation a = -kv^2, where “a” is expressed in m/s^2 and “v” in m/s. The particle starts at x = 0 with a velocity of 9 m/s and when x = 13 m the velocity is found to be 7 m/s. Determine the distance the particle will travel before its velocity drops to 3 m/s and before it comes to rest.

Any help would be appreciated.

Thanks for all input, it was a tremendous help. The book doesn't provide much in the way of examples outside of the standard differentiation/integration to find position, velocity, acceleration.

Ironduke, in this problem k is a constant and my best guess is that the starting angle is not important as it would have been otherwise stated.

Answer 1

Ok here goes, I have done this pretty quickly so please excuse any mistakes,

start with

vdv=adx
note that a=-kv^2, hence you need to find k
now, integrate from v0 to v1 and x0 to x1 i.e.
int(7,9)vdv=int(13,0)(-kv^2dx)
hence
int(7,9)1/v dv=-k*int(13,0)dx
hence
ln(7)-ln(9)=-k*13,
hence k=0.0193

now use the same method for finding distances
i.e.
int(3,9)1/v dv = -0.0193 int(x,0)dx
hence
ln(3)-ln(9)=-0.0193*[x-0]
hence x=56.92 m

in the above read int(a,b) as definite integral from a to b, i.e. a on top of the integral and b on the bottom.

Now if you use the same method for finding the distance when velocity = 0, you will find that x tends to infinity as v tends to 0.

I think this is correct,if you find otherwise let me know too.

Oh btw, in case you needed the proof of the first equality:
v=dx/dt, hence dt=dx/v
a=dv/dt, hence dt=dv/a
hence vdv=adx ..

Answer 2

You do not indicate the angle at which the particle starts. Velocity has two components: speed and direction. You will get a different answer if the particle is started at 15 degrees than if it was started at 45 degrees or straight up at 90 degrees.

Also the units of acceleration are m/s/s and the units of -kv^2 are
m^2/s^2. Thus if the equality is to hold the units of k must be 1/m.
I had the feeling that k was a constant and was dimensionless. What exactly is k?

Answer 3

Treat this exactly as you would a spring problem. The bond itself is the spring. You have decelleration as a result of the tensile resistance of the spring. You can easily find your value of k.....
remember that your atom is not only moving from x=o-->13, but it is also slowing down from 9 m/s to 7 m/s in that distance.

You can find k from that decelleration info as well as distance.

Answer 4

Here is the approach that might help you.

a = -kv^2 = dv/dt
(-1/v^2) dv = k dt......(1)

Assume at x = 0, t = 0.
Integrating (1) gives,
1/v - 1/9 = k t

Solve for v,
v = 1/(k t + 1/9)

x = ∫v dt = ∫1/(k t + 1/9) dt = (1/k)ln(k t + 1/9) - (1/k)ln(1/9)

Can you finish it now?

Answer 5

Find the example in your book. A problem like that is bound to be in there somewhere with different variables.