Question

\[ f(t)=2400 t-600 t^{2} \] denotes the position in miles above the earth of a rocket \( t \) hours after liftoff. Assume that the rocket is always moving in a vertical direction. a) Find the average

Answer 1

velocity of the rocket on the time interval from 1 to 2 hours after launch.

To find the average velocity over a specific interval of time, we need to find the difference in position over this interval, and then divide that by the length of the time interval.

Let's calculate the position at the start and the end of the given time interval.

Step 1:
Calculate the position at 1 hour (t1). We substitute 1 into our position function:

\[ f(1)=2400*1-600*1^{2} \]

Step 2:
Calculate the position at 2 hours (t2), again substituting into our position function:

\[ f(2)=2400*2-600*2^{2} \]

Step 3:
Now that we have our start and end positions, let's calculate the difference in position. This is given by:

\[ \Delta P=f(t2)-f(t1) \]

Step 4:
Next, we calculate the difference in time:

\[ \Delta T=t2-t1 \]

Step 5:
The average velocity Vavg is then calculated by taking the difference in position and dividing it by the difference in time:

\[ V_{avg}=\frac {\Delta P} {\Delta T} \]

So, the average velocity of the rocket on the time interval from 1 to 2 hours after launch is 600.0 miles per hour.