Rates of Change

Question

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, its height in feet after seconds is given by

y = 80t - 16t². Find the average velocity (i.e. the change in distance with respect to the change in time) for the time

period beginning when = 1 and lasting..

a) 0.5 secs

b) 0.1 secs

c) 0.01 secs

d) 0.0001 secs

based on those results in a) to d), guess what the instantaneous velocity of the ball is when t=1?

Answer 1

a) - d) are all just slope calculations using two distinct points on the height curve, so we will use the good old "change in y over change in x" formula: (y₂ - y₁) / (x₂ - x₁) .

For example, for a) we find the height at t = 1.5 and the height at t = 1: y(1.5) = 84 and y(1) = 64 so average velocity is (84 - 64) / (1.5 - 1) = 40 ft/sec

You use the same (x₁ , y₁) for b) - d) , namely (1 , 64).

If you do the calculations correctly, you will see that the avg. velocities in a) - d) are getting arbitrarily close to a whole number, which is what you will register as your guess for the instantaneous velocity at t = 1 sec.