E Paused Time interval Average velocity (m/s) EXAMPLE 3 Suppose that a ball is dropped from the upper observations deck of the CN Tower in Toronto, 450m above the ground. Find the velocity of the ball after 5 seconds. 5Sts6 53.9 5sts 5.1 49.49 SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo's law is expressed by the equation 5Sts 5.05 49.245 5sts 5.01 49.049 5 sts 5.001 49.0049 Video Example) s(t) = 4.9t %3D Tutorial The difficulty in finding the velocity after 5 s 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 interval of a tenth of a second from t = 5 to t = 5.1: Online Textbook change in position average velocity = time elapsed s(5.1) - s(5) 0.1 4.9(5.1 2-4.9( 0.1 The table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that as we shorten the time period, the average velocity is becoming closer to m/s. 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 (instantaneous) velocity after 5
E Paused Time interval Average velocity (m/s) EXAMPLE 3 Suppose that a ball is dropped from the upper observations deck of the CN Tower in Toronto, 450m above the ground. Find the velocity of the ball after 5 seconds. 5Sts6 53.9 5sts 5.1 49.49 SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo's law is expressed by the equation 5Sts 5.05 49.245 5sts 5.01 49.049 5 sts 5.001 49.0049 Video Example) s(t) = 4.9t %3D Tutorial The difficulty in finding the velocity after 5 s 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 interval of a tenth of a second from t = 5 to t = 5.1: Online Textbook change in position average velocity = time elapsed s(5.1) - s(5) 0.1 4.9(5.1 2-4.9( 0.1 The table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that as we shorten the time period, the average velocity is becoming closer to m/s. 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 (instantaneous) velocity after 5
Algebra & Trigonometry with Analytic Geometry
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Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.4: Mathematical Induction
Problem 46E
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Average velocity
Transcribed Image Text:3 Logout Succes x
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Bb General Syllab x
A HW 01 (Sectio x
webassign.net/web/Student/Assignment-Responses/submit?dep=25609279&tags=autosave#question3904441 1
* Cengage Digit x
* Cengage Lear x
+
Time
E Paused
Average
velocity (m/s)
interval
EXAMPLE 3 Suppose that a ball is dropped from the upper observations deck of the CN Tower in
Toronto, 450m above the ground. Find the velocity of the ball after 5 seconds.
5sts6
53.9
5stS 5.1
49.49
SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance
fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for
free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s(t) and measured in
meters, then Galileo's law is expressed by the equation
5sts 5.05
49.245
5Sts 5.01
49.049
5sts 5.001
49.0049
Video Example)
s(t) = 4.9t
Tutorial
The difficulty in finding the velocity after 5 s 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 interval of a tenth of a second from t = 5 to t = 5.1:
Online Textbook
change in position
average velocity =
time elapsed
s(5.1) - s(5)
%3!
0.1
4.9( 5.1
)²-4.9(
0.1
The table shows the results of similar calculations of the average velocity over successively smaller time
periods.
It appears that as we shorten the time period, the average velocity is becoming closer to
m/s. 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 (instantaneous) velocity after 5
s is
m/s
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