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
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