1. Feb 2, 2009

### MC_UNLV

A rocket is fired vertically and tracked by a radar station on the ground, a distance (r) away from the rocket. When the station reads an angle of (theta) = 60* between the rocket and the ground, we are given that the distance r = 30,000ft, r(double-dot) = 70 ft/sec, and theta(dot) = 0.02 rad/sec. Find the magnitude of the velocity and acceleration of the rocket at this position.

I know that to solve this, you need to find r(dot), and that this is somehow related to r as a function of time. I do not understand how to get this relationship, or how to find r(dot). Can anyone please help?

2. Feb 3, 2009

### chrisk

Let

$$\vec{r}=r(\theta{})\hat{r}$$

Differentiating with respect to time and using the chain rule gives

$$\vec{\dot{r}}=\dot{r}\hat{r}+r\frac{d\hat{r}}{d\theta{}}\frac{d\theta{}}{dt}$$

and

$$\frac{d\hat{r}}{d\theta{}}=\hat{\theta}$$

Why? Differentiate this expression again to arrive at an expression for r double dot in terms of the unit vectors r and theta. This should get you started.

3. Feb 3, 2009

### MC_UNLV

Forgive my ignorance, but what does the "^" above r and theta mean, and what is the difference between the r with and without the ^?

4. Feb 3, 2009

### chrisk

The ^ represents the unit vector. In Cartesian coordinates it's

$$\hat{x}\mbox{ and }\hat{y}$$

The r without the hat (^) is the magnitude of r. Have you been exposed to polar coordinates and the associated unit vectors?

5. Feb 3, 2009

### MC_UNLV

Yes, I understand vectors, I just have seen it with different notations.

I still do not understand what you are trying to say with the expressions in your first reply. I do not get how to relate time to the values of r and theta, if a specified time is not given.