Non-Constant Acceleration

[miggitymark]

1. The problem statement, all variables and given/known data

The acceleration of an object in a fluid is proportional to its speed squared. If the object's initial speed is 1.13 m/s, how much time until its speed is reduced by half (0.565 m/s)?

2. Relevant equations

a=-.48v^2

3. The attempt at a solution

I tried using the formula

v(f)-v(0) + a*t

Taking the derivative of the whole thing seems like it's fruitless. If I integrate the formula I can get an equation that relates to velocity:

v=-.16x^3

But I don't know how that relates to the time. Should the "x" be displacement or time since velocity (the result of the equation) is displacement AND time? I can solve for x, but I don't know what I'm actually solving. thanks everyone

[chrisk]

Your equation relating acceleration to velocity can be written as

a=\frac{dv}{dt}=kv^2

Separate the variables and integrate.

[Nabeshin]

I tried using the formula

v(f)-v(0) + a*t



Remember that these quick little equations of motion most of us memorize are all derived from the case of constant acceleration!

[miggitymark]

Ahh, the acceleration IS the derivitive. So the integral with -.16 is the equation I need to use? What is the X in that equation then? time or distance?

[chrisk]

Separating variables gives

\frac{dv}{v^2}=kdt

Integrate both sides using the appropriate limits for v (v0 to v0/2) and t (0 to t).