# Non-Constant Acceleration

1. Jan 30, 2009

### 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

2. Jan 30, 2009

### chrisk

Your equation relating acceleration to velocity can be written as

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

Separate the variables and integrate.

3. Jan 30, 2009

### Nabeshin

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

4. Jan 30, 2009

### 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?

5. Jan 31, 2009

### 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).