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Average Velocity versus Instantaneous Velocity

Given a position function find the average velocity

Average velocity can be found by:

timeinchange

positioninchange

t

smvelav =

==

sec

=)(ts position function [also called )(tx ]

Let 5016)( 2 += tts where =t time in seconds and

( )=ts feet

What is the average velocity on [ 0, 1]

01)0()1(

= ssvelav

1

5034 =

sec16

ft=

Since this is negative, the object is moving either left or ina downwards direction.

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Now, that we know that )(')( tstv = we can easily find the

instantaneous velocity at some value oft

Lets find )1(v [It should be close to our last average

velocity.]

( )

sec32)(

5016)(

)(')(

2

ftttv

tdtdtv

tstv

=

+=

=

So,

sec

32)1(ft

v =

Lets do page 117 #94

A ball is thrown straight down from the top of a 220-foot

building with an initial velocity of -22 feet per second.

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What is its velocity after 3 seconds? What is its velocity

after falling 108 feet?

Given:0

216)( stvttso++=

So for this problem, 2202216)( 2 ++= ttts

We need to first find )(tv

( )

2232)(

2202216)(

)(')(

2

=

+=

=

ttv

ttdtdtv

tstv

Its velocity after 3 seconds will be )3(v

sec11822)3(32)3(ft

v ==

If the ball fell 108 feet, then its position would be 220

108 = 112 ft

We need to find the time when the ball is at 112 ft

Let 112)( =ts and solve for t

2202216112 2 += tt

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10822160 2 += tt Hence, 2=t seconds

Now we just need to find )2(v

sec8622)2(32)2(

ftv ==

What is the balls terminal velocity? [What is its velocity

when it hits the ground?]

First find when it hits the ground by letting 0)( =ts

[Your TI will be handy for this!]

22022160

0)(2

+=

=

tt

ts

t

Terminal velocity is =)(v

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During the first 40 seconds of a rocket flight, the rocket is

propelled straight up so that in t seconds it reaches a

height given by 10)(

3t

ts = feet

(a) What is its position at 40=t seconds?

(b) What is its average velocity on [ 0, 40 ] ?

(c) Find its velocity at 40=t seconds

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See page 117 #98

Time [minutes]

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