Solving Dynamics Problem: Find Velocity, Distance & Rest Point

[ncr7]

Hey I have been looking at the problem for a long time and for some reason I can't think of how to solve for it... I know I have to do some Calc to figure it out.

Ok so here it is

The acceleration of a particle is defined by the relation a = -0.05[v]^{2}, where
a is expressed in m/<s>^{2}</s> and v in m/s. The particle starts at s=0m with a velocity of 5 m/s. Determine (a) the velocity v of the particle after it travels 10 m, (b) the distance s the particle will travel before its velocity drops to 2 m/s, (c) the distance s the particle will
travel before it comes to rest.

so far I think what I have to do is take the derivative of a which ends up being -.1v but after that I am stuck because now I have velocity as a function of velocity... any ideas? I realize I am most likely doing this totally wrong.

 

[Doc Al]

Since a = dv/dt, you'll need to integrate to find v. Start there.

 

[ncr7]

so I'll get
\intdv=\inta dt
(v2-v1) = -0.05v^{2}*t - 0 (since t_{0}=0)

but I now have 2 unknowns though.
v and t

since I know the change in position shouldn't I do an integration over that? I'm just unsure how exactly it should be done

 

[Doc Al]

so I'll get
\intdv=\inta dt
(v2-v1) = -0.05v^{2}*t - 0 (since t_{0}=0)

Before you can integrate properly, you must separate the variables:
\frac{dv}{dt} = -0.05v^2

\frac{dv}{v^2} = -0.05 dt

Now try integrating. (And don't forget the integration constants.)

 

[ncr7]

ok, I'm starting to have some weird mistakes because I have been on this problem to long lol... but I don't have two integration constants for t or v. I have the change in s from 0m to 10m. This would only really be able to give me where it's time when it comes to a complete stop which is part of the problem. But right now I need to find velocity at 10m.

 

[Doc Al]

One step at a time. First complete the integration to find v as a function of t. Then, using v = dx/dt, integrate once more.

 

[Shooting Star]

In this case, it would save a lot of trouble if he writes a = v(dv/dx). All the answers want some x at some v.

 

[Doc Al]

Even quicker! (I was trying to keep it as straightforward and simple as possible.)

 

[ncr7]

ok I think I got it.

so taking

\int_0 ^{10m} -0.05v^2 dx = \int_0^v v dv

I get-0.05x|_0 ^{10} = ln|v||_{5m/s} ^v

e^{1.11} = v

getting 3.03 m/s

I think that's right...

 

[Doc Al]

Looks good, but be careful how your write up your work.
v\frac{dv}{dx} = -0.05v^2

\frac{dv}{v} = -0.05dx

ln(v) = -0.05x + C

 

[ncr7]

ok yeah I'm going to write it more throughly when I put it on paper. There wouldn't be the C because I have known points I'm integrating between. thanks for checking my work.