Xva‐vs‐t_BASICS See Page 2 for v = slope of x vs t and a = slope of v vs t See Page 3 for area relationships Pictorial Time Lapse Examples of 1 Dimensional Translational ( Linear) Motion Each picture above shows a car traveling along a straight path. Its position is shown at equal time intervals with a clock above it displaying the elapsing time. Graphs of the linear displacement versus time (x vs t), velocity versus time (v vs t) and acceleration versus time (a vs t) are shown for each situation. In 1 to 3 the car travels with three different constant velocities, slowest in 1 and fastest in 3. In 4 the car starts from rest and travels at a constant positive acceleration. In 5 the car travels a constant negative acceleration. Note the graphs, and the velocity (black) and acceleration (red) vectors. They are above the car and under the clock. They are very small to use high zoom value, like 200%. 45
15
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4
5
0
0
0
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x
v
4.0
3.0
2.0
1.0
0.25
0.50
0.75 1.00
time
a
a cce le ra tio n
0
veloci ty
2
0
displ acement
1.
1
t
0.25
0.50
t 0
0.75 1.00
time
0 .2 5
0 .5 0
t
0 .7 51 .0 0
time
45
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30
30
5
0
0
45
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45
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x
30
30
v
4.0
3.0
2.0
1.0
0.25
0.50
t
0.75 1.00
time
2
1
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0
3.
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0.75 1.00
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time
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a
0 .2 5
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0 .7 5 1 .0 0
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x
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0 .5 0
acc el era tio n
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acc el er atio n
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v
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3.0
2.0
ve locity
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di sp laceme nt
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ve locity
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1
displ acement
1.0
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0.50 0.75 1.00
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0.75 1.00
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Page 1 of 3 t 0
0.75 1.00
a
v
4.0
3.0
2.0
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0.25
0.50
0.75 1.00
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0.50
time
a
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0 .5 0
0 .7 51 .0 0
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x
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veloci ty
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d ispla ce ment
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Xva‐vs‐t_BASICS The picture below show the graphs for a constant velocity motion for 5 seconds from time t = 0 sec to t = 5 secs. During this time the body travels from displacement x= 0 meters to x = 10 meters. The first graph shows the displacement versus time, x vs t, graph. The second graph shows the resulting velocity versus time , v vs t, graph. The third graph shows the resulting acceleration versus time , a vs t, graph. Both display definitions and calculation. m (meters)
x
(displacement)
(5,10)
10
8
6
4
2
x=10m
(0,0) t=5s
0 1 2 3 4 5 6 7 8 10
s (seconds)
t (time)
m/s (meters/sec)
v
(velocity)
velocity v =
10
8
6
4
2
x t
= change in displacement x change in time t
t=5s v=0 m/s
0 1 2 3 4 5 6 7 8 10
s (seconds)
t (time)
= slope of x vs t graph
= slope of displacement versus time graph 10m
=2m/s
=
5s
m/s2 (meters/sec2)
a
aceleration a = v t
(acceleration)
10
8
= change in velocity v change in time t
6
4
2
0 1 2 3 4 5 6 7 8 10
s (seconds)
t (time)
= slope of v vs t graph
= slope of velocity versus time graph
0 m/s =0 m/s 2
= 5s
Page 2 of 3 Xva‐vs‐t_BASICS Demonstration that x = area under the v vs t graph ‘ change in displacement = area under the velocity versus time graph’ And v = area under the a vs t graph ‘ change in velocity = area under the acceleration versus time graph’ x
x0
1
v
x’
x0
t
v
v0
t
x = x’- x
= v0 t
t t’
= v0 t’
1
v0
 x = v0 t
= area under a
v vs t graph
t
a
a
t’ t
Page 3 of 3 v’
v0
a
t
t
v
v0
v
v
x
a
t
v = v’- v
=at
t’ t = a t’
v = a  t
= area under
a vs t graph
t’ t