Calculating Distance Traveled from Acceleration and Velocity Relation

[gnarly]

Homework Statement

The acceleration of a particle is defined by the relation

a=-k(v)^2 k is a constant.

The particle starts at x=0 with a velocity of 20 m/s. At x=100 m its velocity is 15 m/s

Determine the distance traveled by the particle (starting from x=0)
A) before its velocity drops to 10 m/s

B)before it comes to rest.



This is the only problem in college I encountered that I couldn't solve. I took calculus 3 it seems to be so simple yet I have no idea how to solve. I don't get how I can find the position without the time traveled. I don't get how acceleration can be dependent on the velocity. I spent so many hours tryying to solve this. It seems like i could integrate a twice to obtain the position function but that turns out to be wrong. I can't believe that math class seems so easy to me but i can't even solve the math problems in a standard first year physics class. This seems to be a differential equations problem but it can't be as its in a first year physics class. Sorry i couldn't really attempt a solution because i don't know where to begin. Any help will be greatly appreciated.

 

[Dick]

a=dv/dt. So you've got dv/dt=(-kv^2). That's a separable ODE. Solve for v first. Then integrate v to get the position.

 

[gneill]

So you're given that

$$a = -k*v^2$$

but

$$a = \frac{dv}{dt}$$

$$\frac{dv}{dt} = -k*v^2$$

$$\frac{dv}{v^2} = -k dt$$

etc.

When you integrate, make sure to incorporate your as one of the limits.

You'll have to pull this trick again to find the distance expression. After that it's just mixing and matching the formulae and given values to solve for k, then on to the particular cases.