### Exercises for the Velocity Module

• 1:
Compute the average velocity of the canteloupe over the time interval 2.0  t  < 2.5 and over the interval 0.5  t  < 1.5.

• 2:
What is the average velocity of the canteloupe over the interval [0.99,1]? ... over the interval [1,1.01]? Use the following table to get average velocities over correspondingly shorter intervals:  t(secs) 0.999 1 1.001 x(feet) 84.027 84 83.973
What value (to two decimal places, say) should we assign to the instantaneous velocity of the canteloupe at time t = 1 sec?

• 3 :
• a.) Even if you did the above problems by hand, why not use Maple this time? Approximate the instantaneous velocity of the canteloupe at time t=2 sec by first defining the function
 f:= t -> 95 + 5*t - 16*t^2;
and then calculating average velocities over short intervals which begin or end at t=2:
       (f(2.01) - f(2))/ (2.01-2);
       (f(2)-f(1.99))/(2-1.99);
and so forth. What value would you assign for the instantaneous velocity at t=2?
• b.) You can get fancier: Let's define the function
      vav := t -> (f(t+h)-f(t))/h;
This will give the average velocity over the time interval [t,t+h] once we have assigned a value to h. Do part (a) again using vav. Then use vav to find the instantaneous velocity at several (at least three) other values of t.
• c.) It's even better if you graph it. Assign a (small) value to h, and then
        plot({f(t),vav(t)},t=0..2.6)
Do this several times with decreasing values of h (but only turn in \it one plot). What do the graphs of all the 'vav' s have in common? How do they differ? How is the graph of f(t) related to that of vav(t) . In particular how does f behave when vav is positive? negative? big? small? -- Is this behavior precise or only approximate?

• 4:
• a.) Use Maple's solve command to determine precisely at what time the canteloupe hits the ground.
• b.) How would you determine the highest height reached by the canteloupe? At what time does this occur?