Solving Dynamics Problem: Find Velocity, Distance & Rest Point

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In summary, the conversation discusses solving for the acceleration of a particle given its velocity and distance traveled. The method involves taking the derivative of a function and integrating to find the velocity at a given distance. The conversation also touches on the importance of separating variables in order to properly integrate.
  • #1
ncr7
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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[tex][v]^{2}[/tex], where
a is expressed in m/[tex]^{2}[/tex] 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.
 
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  • #2
Since [itex]a = dv/dt[/itex], you'll need to integrate to find v. Start there.
 
  • #3
so I'll get
[tex]\int[/tex]dv=[tex]\int[/tex]a dt
(v2-v1) = -0.05[tex]v^{2}[/tex]*t - 0 (since [tex]t_{0}[/tex]=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
 
  • #4
ncr7 said:
so I'll get
[tex]\int[/tex]dv=[tex]\int[/tex]a dt
(v2-v1) = -0.05[tex]v^{2}[/tex]*t - 0 (since [tex]t_{0}[/tex]=0)
:yuck:

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

[tex] \frac{dv}{v^2} = -0.05 dt[/tex]

Now try integrating. (And don't forget the integration constants.)
 
  • #5
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.
 
  • #6
One step at a time. :wink: First complete the integration to find v as a function of t. Then, using v = dx/dt, integrate once more.
 
  • #7
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.
 
  • #8
Even quicker! :smile: (I was trying to keep it as straightforward and simple as possible.)
 
  • #9
ok I think I got it.

so taking

[tex] \int_0 ^{10m} -0.05v^2 dx = \int_0^v v dv [/tex]

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

[tex] e^{1.11} = v [/tex]

getting 3.03 m/s

I think that's right...
 
Last edited:
  • #10
Looks good, but be careful how your write up your work.
[tex] v\frac{dv}{dx} = -0.05v^2[/tex]

[tex] \frac{dv}{v} = -0.05dx[/tex]

[tex] ln(v) = -0.05x + C[/tex]
 
  • #11
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.
 

1. What is the first step in solving a dynamics problem?

The first step in solving a dynamics problem is to identify the problem and its given parameters. This includes determining the initial and final states of the system, as well as any external forces acting on the system.

2. How do I find the velocity of an object in a dynamics problem?

To find the velocity of an object in a dynamics problem, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. You can also use the formula v^2 = u^2 + 2ad, where d is the distance traveled.

3. What is the rest point in a dynamics problem?

The rest point in a dynamics problem is the point at which the velocity of the object becomes zero. This can occur when the object reaches its maximum height in a projectile motion problem, or when it comes to a complete stop in a motion with friction problem.

4. How do I find the distance traveled in a dynamics problem?

To find the distance traveled in a dynamics problem, you can use the formula d = ut + 1/2at^2, where d is the distance, u is the initial velocity, a is the acceleration, and t is the time. You can also use the formula d = v^2 - u^2 / 2a, where v is the final velocity.

5. Can I use the same equations to solve dynamics problems with different types of motion?

Yes, the same equations can be used to solve dynamics problems with different types of motion, such as projectile motion, circular motion, or motion with friction. However, the given parameters and initial conditions may vary, so it is important to properly identify and apply the correct equations for each specific problem.

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