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e 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 (-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-tor-5.1. average velocity change in position time lapsed 8051)-83) 1

e 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 (-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-tor-5.1. average velocity change in position time lapsed 8051)-83) 1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 tot - 5.1. change in position average velocity passic.com time clapsed Time interval V- 5sts6 m/s. The following table shows the results of similar calculations of the average velocity over successively smaller time periods. 5sts 5.1 5sts 5.05 - 5sts 5.01 5sts 5.001 s(5.1)-(5) 0.1 4.9 Average velocity (m/s) 53.9 49.245 49.049 0.1 49.0049 It appears that as we shorten the time period, the average velocity is becoming closer to m/s m/s (rounded to one decimal place). The instantaneous velocity when - 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 (instantaneous) velodity after 5s is the following. (Round your answer to one decimal place.)
Transcribed Image Text: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 tot - 5.1. change in position average velocity passic.com time clapsed Time interval V- 5sts6 m/s. The following table shows the results of similar calculations of the average velocity over successively smaller time periods. 5sts 5.1 5sts 5.05 - 5sts 5.01 5sts 5.001 s(5.1)-(5) 0.1 4.9 Average velocity (m/s) 53.9 49.245 49.049 0.1 49.0049 It appears that as we shorten the time period, the average velocity is becoming closer to m/s m/s (rounded to one decimal place). The instantaneous velocity when - 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 (instantaneous) velodity after 5s is the following. (Round your answer to one decimal place.)
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