# Homework Help: Kinematics problem - struggling with my calculus

1. Apr 21, 2012

### Femme_physics

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

http://img62.imageshack.us/img62/6705/smallproj.jpg [Broken]

3. The attempt at a solution

Basically I'm struggling to understand what am I supposed to differentiate. If it's V, which is 60, then the derivative of a number is zero. If it's a, then, a = -1.2v2

If I am supposed to use a = dv/dt , which I believe I do because it relates everything I need...then I could use a little guidance and help in my calculus.

http://img51.imageshack.us/img51/7691/dvdtq.jpg [Broken]

Last edited by a moderator: May 5, 2017
2. Apr 21, 2012

### ehild

The acceleration is a=dv/dt=-0.4v^3. dv/dt=-0.4v^3 is a differential equation to be solved for v(t) with the initial condition v(0)=60 m/s.
You have studied solving differential equations, I guess.

ehild

3. Apr 21, 2012

### Femme_physics

Last edited by a moderator: May 5, 2017
4. Apr 21, 2012

### ehild

No,Vi= 60 m/s is the initial velocity, and you need to find V(t) the velocity as function of time.

Are you familiar with differential equations? dv/dt=-0.4v^3 is a first-order linear separable one. Rearrange the equation

$$\frac{dv}{v^3}=-0.4dt$$,

and integrate both sides. The left side is integrated from v(0)=60 to v(t), the right side goes from t=0 to t.
$$\int_{60}^{v(t)}{\frac{dv}{v^3}}=\int_0^t{-0.4d\tau}$$

ehild

5. Apr 21, 2012

### Femme_physics

I am contemplating about this exercise. I realize we're dealing with related rates of change. As velocity changes so does acceleration. Hence calculus must be used because acceleration is not constant and is a function of velocity.

I think the graph looks something like this:
http://img696.imageshack.us/img696/1498/timechange.jpg [Broken]

We're taking the derivative of velocity with respect to time, because this tells us how the function changes with respect to time, therefor we can use it to elicit what the velocity is at a specific time via the formula a = dv/dt , if we integrate.

OK, then, I think I got the theory. As far as the actual math, I'm not experienced in calculus-- so I'll ask my friend today and I will try to get to the solution with her. I'll come back to this thread once I got the math straightened out :)

Last edited by a moderator: May 5, 2017
6. Apr 23, 2012

### Femme_physics

Last edited by a moderator: May 5, 2017
7. Apr 23, 2012

### ehild

If it is V(t), it does not look like the real one. Try to plot V(t) you have got.Or see picture.

Congratulation!!!!

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8. Apr 23, 2012

### Femme_physics

Thanks :)

And yes, your graph makes sense but I thought that if V is going downwards I use a minus sign, but on second thought minus is a property of how acceleration behaves on the velocity, not velocity itself. Maybe it would be wiser to leave the leading direction of velocity (down in this case) as plus, always?

9. Apr 23, 2012

### ehild

Yes, that rocket can not turn upward, can it? So let the velocity be plus.
The 60 m/s for initial value meant downward velocity, into the water, not out of it.
The acceleration was negative with respect to the velocity (deceleration).

ehild