Example 3 Video Example) Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds. Solution The difficulty in finding the velocity after 5 seconds is that we are dealing with a single instant of time (t = 5), so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second from t = 5 to t = 5.1. average velocity = - change in position time elapsed s(5.1) - 5(5) 0.1 4.9 - - 4.9 0.1 m/s. The following table shows the results of similar calculations of the average velocity over successively smaller time periods. Time Average interval velocity (m/s) 5 sts 6 53.9 5 sts 5.1 49.49 5 s t s 5.05 49.245 5 s t s 5.01 49.049 5 s t s 5.001 49.0049 It appears that as we shorten the time period, the average velocity is becoming closer to (instantaneous) velocity after 5 s is the following. (Round your answer to one decimal place.) m/s (rounded to one decimal place). The instantaneous velocity when t = 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t = 5. Thus the V = m/s Need Help? Read It