PHYS 120 – Lecture 2
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PHYS 120
Today's Concepts:
Dr. Bernd Stelzer
a) Position, Velocity, Acceleration
b) 1-D Kinematics, graphs and constant acceleration
Pre-Lecture Thoughts
Are there topics in this section that you still find confusing or that
need further explanation?
“I need help on the graphs, how to interpret the relationship between
each of the three quantities.”
“The idea behind solving question #2 was still a bit confusing,
especially why it couldn’t be 2ft travelled between t=1s and 2s.”
“Not so far. Although I haven't learned anti derivatives and integrals
in Calculus II yet, but I will hopefully learn more about that before we
require it in this course.”
“I did not really find anything that confusing during this section of
physics, I overall found it as a nice review to physics 12 since I did
forget a lot of things.”
1D Position vs Time Graphs
vavg =
x
t
Displacement – change in position: Δx = x2 – x1
Time elapsed – change in time: Δt = t2 – t1
x
Slope of the
line drawn
through the
X2
two points
X1
t1
t2
t
Position vs Time Graphs
vavg =
x
t
Displacement
Time elapsed
Decrease the time interval
x
t2
t1
t
Position vs Time Graphs
Average Velocity
vavg =
x
t
Displacement
Time elapsed
Decrease the time interval
x
t1
t2
t
1D Position vs Time Graphs
vavg =
lim
x
t
Decrease the time interval
x
t!0
dx
v=
dt
Instantaneous
velocity = time
derivative of the
position = the
slope of the
tangent line at
time t
t
Slope = v
t
Differential Calculus
Differentiation (taking a derivative) is a convenient method for
calculating the constantly changing slope of a nonlinear curve.
Consider a function
where C and n are constants
The first derivative of y with respect to x is:
Example
Discussion Question
Which instantaneous velocity-vs-time graph goes with the
position-vs-time graph on the left?
x
v
v
v
v
The v(t) vs. t graph is the slope of the x(t) vs. t graph.
Slope begins with a constant positive value and later decreases to zero.
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10
Acceleration vs Time Graphs
Slope = 0
Slope < 0
Slope > 0
Position
a(t) is the slope of
the v(t) vs. t graph
a=0
a>0
a<0
PHYS120
11
Bridge Question: x(t), v(t), a(t) plots
For the Position and Velocity curves shown on the left,
which is the correct plot of acceleration vs. time?
Position
A
negative
slope
positive
slope
B
Position from Velocity vs Time Graph
A velocity-vs-time graph can be made by finding the slope of
the position-vs-time graph.
slope
v
x
v
area
A position-vs-time graph can be made by finding the area
under the velocity-vs-time graph. – Let’s see how this works.
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13
Position from Velocity vs Time Graph
x (m)
v (m/s)
Constant velocity
50
40
30
20
10
0
=vavg Δt = Δx
Area = 30 m/s x 1 s = 30 m
0
1
2
3
4
t (s)
5
0
1
2
3
4
t (s)
5
180
150
120
90
60
30
0
Assuming x = 0 at t = 0
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14
Position from Velocity vs Time Graph
x (m)
v (m/s)
Constant velocity
50
40
30
20
10
0
Area = 30 m/s x 2 s = 60 m
0
1
2
3
4
t (s)
5
0
1
2
3
4
t (s)
5
180
150
120
90
60
30
0
PHYS120
15
Position from Velocity vs Time Graph
x (m)
v (m/s)
Constant velocity
50
40
30
20
10
0
Area = 30 m/s x 3 s = 90 m
0
1
2
3
4
t (s)
5
0
1
2
3
4
t (s)
5
180
150
120
90
60
30
0
PHYS120
16
Position from Velocity vs Time Graph
x (m)
v (m/s)
Constant velocity
50
40
30
20
10
0
Area = 30 m/s x 4 s = 120 m
0
1
2
3
4
t (s)
5
0
1
2
3
4
t (s)
5
180
150
120
90
60
30
0
PHYS120
17
Position from Velocity vs Time Graph
v (m/s)
Constant velocity
50
40
30
20
10
0
Area = 30 m/s x 5 s = 150 m
x (m)
0
1
180
150
120
90
60
30
0
2
3
4
t (s)
5
3
4
t (s)
5
x = 30 t m
=v
0
1
t
2
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Position from Velocity vs Time Graph
Linear increasing velocity
v (m/s)
60
40
20
0
0
2
3
4
t (s)
5
1
Area = (10[m/s] ⇥ 1[s]) = 5m =vavg Δt = Δx
2
150
x (m)
1
100
50
0
0
1
2
3
4
PHYS120
t (s)
5
19
Constant Acceleration
Linear increasing velocity
v (m/s)
60
40
20
0
0
2
3
4
t (s)
5
1
Area = (20[m/s] ⇥ 2[s]) = 20m
2
150
x (m)
1
100
50
0
0
1
2
3
4
PHYS120
t (s)
5
20
Constant Acceleration
Linear increasing velocity
v (m/s)
60
40
20
0
0
2
3
4
t (s)
5
1
Area = (30[m/s] ⇥ 3[s]) = 45m
2
150
x (m)
1
100
50
0
0
1
2
3
4
PHYS120
t (s)
5
21
Constant Acceleration
Linear increasing velocity
v (m/s)
60
40
20
0
0
2
3
4
t (s)
5
1
Area = (40[m/s] ⇥ 4[s]) = 80m
2
150
x (m)
1
100
50
0
0
1
2
3
4
PHYS120
t (s)
5
22
Constant Acceleration
Linear increasing velocity
v (m/s)
60
40
20
0
0
2
3
4
t (s)
5
1
Area = (50[m/s] ⇥ 5[s]) = 125m
2
150
x (m)
1
100
50
0
0
1
2
3
4
PHYS120
t (s)
5
23
Constant Acceleration
Linear increasing velocity
v (m/s)
60
40
v=a
t
20
0
0
1
2
3
4
2
3
4
t (s)
5
x (m)
150
1
x= a
2
100
50
t2
0
0
1
PHYS120
t (s)
5
24
Constant Acceleration
Linear increasing velocity
v (m/s)
60
40
20
0
0
1
2
3
4
3
4
t (s)
5
x (m)
150
100
slope increasing
50
0
0
1
2
PHYS120
t (s)
5
25
Position from Velocity vs Time Graph
(vx is constant and >0)
(vx is constant and <0)
(vx varies with time)
General v(t) vs. t curve.
• The total displacement from t1 to t2 is the area
under the curve for this interval.
• This can be calculated by summing the areas
of (infinitesimal) rectangles. (integral)
N
Δx = x f − xi = lim ∑ vx Δt =
Δt→0
i=1
tf
∫ vx dt
ti
Don’t
Panic!
Discussion Question
Which position-vs-time graph goes with the velocity-vs-time graph at
the top? The particle’s position at t = 0 s is x0 = –10 m.
Area > 0
Area < 0
dx/dt > 0
dx/dt < 0
Velocity from Acceleration vs Time Graph
Constant acceleration
a (m/s2)
15
10
5
Similarly, the area under a constant acceleration-vs-time
graph gives a linear velocity-vs-time graph
0
0
1
2
3
4
t (s)
5
3
4
t (s)
5
v (m/s)
60
40
vf = vi + a
t
20
0
0
1
2
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Kinematic Equations
In general:
tf
ti
constant
a(t) = a
Throwing a ball vertically – free fall motion
x
vx > 0
vx < 0
Constant negative
acceleration: ax= -9.81 m/s2
𝑣⃗
x=0
Slowing down
as it rises:
vx is positive and
becoming smaller
x(t),m
Speeding up
going down:
vx is negative
and becoming
more negative
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