Distance, Displacement, and Acceleration for Linear Motion Lesson 10-2.

Distance, Displacement, and Acceleration for Linear Motion

Lesson 10-2

Distance vs. Displacement

5 steps

Distance = 5 stepsDisplacement = 5

steps

2 steps

Distance = 7 stepsDisplacement = 3 steps

5 steps

Distance = 12 stepsDisplacement = –2 steps

Displacement can be negative when you are on the other side of the starting point.

Distance, however, is always non-negative.

Definitions

Displacement velocityb

a

dt

Distance velocityb

a

dtAbsolute value ensures distance is non-negative.

Example 1:A moving object has

feet per second in the time interval [0, 3].

find the time subintervals in which the

velocity is positive and in which it is negative.

3 2( ) 2 2v t t t t

3 22 2 0t t t 1 1 2 0t t t

1, 1, 2t t t 1, 1, 2t t t Outsid

e domain

t = 1 is a critical number

Subintervals

[0,1) negative

(1,3] positive

Example 1:A moving object has

feet per second in the time interval [0, 3].

find the time subintervals in which the

velocity is positive and in which it is negative.

3 2( ) 2 2v t t t t

t = 1 is a critical number

Subintervals

[0,1) negative

(1,3] positive

Example 1:Find the displacement of the object over

the interval.

3

3 2

0

Disp 2 2t t t dt 3

4 3 2

0

1 2 12

4 3 2 t t t t

33 2

0

Dist 2 2t t t dt

30.916 ft

Find the distance of the object over the interval.

27.75 ft

0 3

3 2 3 2

1 1

2 2 2 2t t t dt t t t dt

Example 2:An object has an acceleration given by

(ft/sec/sec) on the interval.

[0, 2π] with an initial velocity of –2. Find

its displacement and distance on the interval.

( ) 2cosv t t dt 2sin t c 2 2sin(0) c

( ) 2cosa t t

Since initial velocity is –2 2c

2

0

disp 2sin 2t dt

2

0

2cos 2t t

12.566 ft

Example 2:An object has an acceleration given by

(ft/sec/sec) on the interval.

[0, 2π] with an initial velocity of –2. Find

its displacement and distance on the interval.

( ) 2cosa t t

2

0

dist 2sin 2t dt

12.566 ft

0

2

2sin 2t dt

Example 3:An object travels along a straight line with

with constant acceleration of 2 ft/sec2.At , the object’s velocity is 11 ft/sec. How far does the object travel during the time

interval when its velocity increases from 15 ft/sec to 21 ft/sec.

( ) 2 1 ft/secv t t

so, ( ) 2v t t c 2( ) 2 ft/ seca t

5sect

therefore,

1c so, 11 2(5) c (5) 11v

Example 3:An object travels along a straight line with

with constant acceleration of 2 ft/sec2.At , the object’s velocity is 11 ft/sec.

How far does the object travel during the time interval when its velocity increases from

15 ft/sec to 21 ft/sec.

( ) 2 1v t t so, 15 2 1t

10

7

dist 2 1t dt

5sect

Since we don’t know the time interval when the velocity increases we must find it so as to know the limits of integration.

7t

and, 21 2 1t 10t

54ft

If the car’s initial velocity was 20 mi/h approximately how far did the car go in this 24 second interval?

Average Acceleration = ½(1.3+1.7)Average Acceleration = ½(1.7+2.2)

Make sure to pay attention to your units