chap. 2: motion in 1d part ipeople.physics.tamu.edu/kamon/teaching/phys218/slide/2013a/lec0… ·...
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Kinematics (1D)
1. Review of Derivative & Integrals
2. Translational Motion (1D) with ISEE
(i) Displacement, Velocity and Acceleration
(x-t, v-t, and a-t graphs)
(ii) Kinematic Equations
Chap. 2: Motion in 1D
Part I
http://people.physics.tamu.edu/kamon/teaching/phys218/
1
Kinematics (1D)
Calculus 1
1) Very Short Review of “derivative” and “integral”
(a) taking limit
(b) derivative method
2) For Exam 1 (Chap. 1-3), I expect you will be familiar with “calculus 1.” (see slides at my phys218 web site.
2
http://people.physics.tamu.edu/kamon/teaching/phys218/
Kinematics (1D)
Review: Integral (I)
& Derivative of I
nnn
nn
n
tAtnAn
CAtntt
I
tI
nCtAn
ttAttf
tAtfI
1 1
1
1
1
d
d
d
d
orespect t with of Derivative
1)( 1
1d d )(
)( of Integral
1)1(1
1
f(t)
f(t4)
1 2 3 4 5
Calculate the area of each strip
∫um the area of each strip
Unknown constant
Area 4 = f(t4) x (t5-t4)
3
t4
Kinematics (1D)
1144
5
1
1013
fi
3
1313
1
5
13
fi
3
1313
5
1
13
fi
3
1
fi
1757154
155
4
1
710
15
13
1 d
1 to 5 from 7 5 )( of Integral (3)
5513
115
13
15
13
1 d
1 to 5 from 5 )( of Integral (2)
1513
155
13
15
13
1 d
5 to 1 from 5 )( of Integral (1)
1
1d to from )( of Integral
f
i
f
i
f
i
f
i
f
i
f
i
f
i
f
i
T
T
T
T
n
T
T
T
T
n
T
T
X
X
n
T
T
n
T
T
nn
ttttAI
T Txtf
tttAI
T Tttf
tttAI
T Tttf
.,g.e
Atn
ttATtTt tAxf
4
Kinematics (1D)
Translational Motion (1D)
1. Treat an object as a particle
2. Concepts of:
Displacement: Dx [ Change in Position ]
Velocity: v [ Rate Change in Displacement ]
Acceleration: a [ Rate Change in Velocity ]
a) Motion with zero acceleration
b) Motion with non-zero acceleration
3. x-t, v-t, a-t graphs
Let’s practice those concepts and calculus
7
Kinematics (1D)
Average Velocity and Instantaneous Velocity
8
Kinematics (1D)
TWO 1D Motions
Let’s take a look at two kinds of 1D
motion. The equations of the motions are:
33
22
)m/s (0.156
1 m/s) (1.0)(
)m/s 0.5(2
1m) (2.0)(
ttty
ttx
9
Kinematics (1D)
x(t) = x0 + v0 t + (1/2) a0 t 2
v(t) = v0 + a0 t
a(t) = a0
Calculus and Eqs. of Motion [I]
Kinematic Eqs. are related by derivatives and integrals.
Motion with constant acceleration.
Initial velocity
Initial position
10
Kinematics (1D)
x(t) = x0 + v0 t + (1/2) a0 t 2 + (1/6) a1 t
3
v(t) = v0 + a0 t + (1/2) a1 t 2
a(t) = a0 + a1 t
Calculus and Eqs. of Motion [II]
Kinematic Eqs. are related by derivatives and integrals.
Motion with varying acceleration.
Initial velocity
Initial position
Extra term!
11
Kinematics (1D)
x(t) = (2.0 m) – (1/2) (0.50 m/s2) t 2
vx(t) = – (0.50 m/s2) t
ax(t) = – 0.50 m/s2
Motion along with x axis
Kinematic Eqs. are related by derivatives and integrals.
Motion with constant acceleration.
Initial velocity = zero
Initial position = 2.0 m
[Eq. 1]
Example 1
12
Kinematics (1D)
y(t) = (1.0 m/s) t + (1/6) (0.150 m/s3) t 3
vy(t) = (1.0 m/s) + (1/2) (0.150 m/s3) t 2
ay(t) = (0.150 m/s3) t
Motion along with y axis
Motion with varying acceleration.
Initial velocity = 1 m/s
Initial position = zero
[Eq. 2]
Kinematic Eqs. are related by derivatives and integrals.
Example 2
13
Kinematics (1D)
TWO 1D Motions (Eqs. 1 & 2)
Study each motion between 0 s and 10 s
using spreadsheet:
Calculate position every 0.5 s and plot them.
Calculate displacement
Calculate average velocity
Calculate velocity every 0.5 s
Calculate acceleration every 0.5 s
14
Kinematics (1D)
1D Motion along with x axis 1D Moion along with y axis
time (s) x (m) dx dx/dt vx ax y (m) dy dy/dt vy ay
0.00 2.00 -0.06 -0.13 0.000 -0.500 0.00 0.503 1.006 1.000 0.000
0.50 1.94 -0.19 -0.38 -0.250 -0.500 0.50 0.522 1.044 1.019 0.075
1.00 1.75 -0.31 -0.63 -0.500 -0.500 1.03 0.559 1.119 1.075 0.150
1.50 1.44 -0.44 -0.88 -0.750 -0.500 1.58 0.616 1.231 1.169 0.225
2.00 1.00 -0.56 -1.13 -1.000 -0.500 2.20 0.691 1.381 1.300 0.300
2.50 0.44 -0.69 -1.38 -1.250 -0.500 2.89 0.784 1.569 1.469 0.375
3.00 -0.25 -0.81 -1.63 -1.500 -0.500 3.68 0.897 1.794 1.675 0.450
3.50 -1.06 -0.94 -1.88 -1.750 -0.500 4.57 1.028 2.056 1.919 0.525
4.00 -2.00 -1.06 -2.13 -2.000 -0.500 5.60 1.178 2.356 2.200 0.600
4.50 -3.06 -1.19 -2.38 -2.250 -0.500 6.78 1.347 2.694 2.519 0.675
5.00 -4.25 -1.31 -2.63 -2.500 -0.500 8.13 1.534 3.069 2.875 0.750
5.50 -5.56 -1.44 -2.88 -2.750 -0.500 9.66 1.741 3.481 3.269 0.825
6.00 -7.00 -1.56 -3.13 -3.000 -0.500 11.40 1.966 3.931 3.700 0.900
6.50 -8.56 -1.69 -3.38 -3.250 -0.500 13.37 2.209 4.419 4.169 0.975
7.00 -10.25 -1.81 -3.63 -3.500 -0.500 15.58 2.472 4.944 4.675 1.050
7.50 -12.06 -1.94 -3.88 -3.750 -0.500 18.05 2.753 5.506 5.219 1.125
8.00 -14.00 -2.06 -4.13 -4.000 -0.500 20.80 3.053 6.106 5.800 1.200
8.50 -16.06 -2.19 -4.38 -4.250 -0.500 23.85 3.372 6.744 6.419 1.275
9.00 -18.25 -2.31 -4.63 -4.500 -0.500 27.23 3.709 7.419 7.075 1.350
9.50 -20.56 -2.44 -4.88 -4.750 -0.500 30.93 4.066 8.131 7.769 1.425
10.00 -23.00 -2.56 -5.13 -5.000 -0.500 35.00 4.441 8.881 8.500 1.500
15
Kinematics (1D)
Position vs Time
-30.00
-25.00
-20.00
-15.00
-10.00
-5.00
0.00
5.00
0 1 2 3 4 5 6 7 8 9 10 11
t (s)
x(m
)
Position vs Time
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0 1 2 3 4 5 6 7 8 9 10 11
t (s)
y(m
)Sketch the slope vs. time
Sketch the slope vs. time
16
Kinematics (1D)
Velocity vs Time
-6.000
-5.000
-4.000
-3.000
-2.000
-1.000
0.000
0 1 2 3 4 5 6 7 8 9 10 11
t (s)
vx(m
)
Velocity vs Time
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
0 1 2 3 4 5 6 7 8 9 10 11
t (s)
vy(m
)
17
vx(t) = – (0.50 m/s2) t
vy(t) = (1.0 m/s) + (1/2) (0.150 m/s3) t 2
Kinematics (1D)
a-t graph ?
v-t graph ?
18
Kinematics (1D)
Motion with Constant Acceleration
A B C
vy
t
y vy-t graph
Diagnostic Test
21
Kinematics (1D)
A B C
ay
t
y ay-t graph
Diagnostic Test
Motion with Constant Acceleration
22
Kinematics (1D)
A B C
ay
t
y
ay-t graph
Motion with Constant Acceleration
Diagnostic Test
23
Kinematics (1D)
1. Recap: ISEE
1 D Motion & Kinematic Equations
2. More Examples
Chap. 2: Motion in 1D
Part II
24
Kinematics (1D) 25
y
Kinematic Eqs. are related by derivatives and integrals.
Kinematics (1D)
x-t,
Motion with Constant Acceleration
Derivative
Derivative
x-t, v-t, x-t, v-t, a-t graphs
26
Kinematics (1D)
a(t) = (4.2 m/s2)
v(t) = 0 + 0 + (4.2 m/s2) t
x(t) = (2.8 m) + (0 m/s) t + (1/2) (4.2 m/s2) t 2
Math: Derivative and Integrals
27
Kinematic Eqs. are related by derivatives and integrals.
Kinematics (1D)
Answer: You can say v = v0 + a t. However, the velocity can be 2-D (see Chap. 3). Thus, they are more generic formulas. If you are only dealing with an 1-D motion along the y axis, you use vy (or simply v). I am simply preparing for 2-D motion in Chap. 3. Namely, 2-D motion can be treated as a combination of two 1-D motions.
Question from students:
The formula that you gave in class was
vy = v0y + ay t.
Why I can't just say v = v0 + a t ?
28
Kinematics (1D)
x = x0 + v0x t + ½ ax t 2 (1)
vx = v0x + ax t (2) vx
2 = v0x2 + 2ax (x – x0) (3)
Kinematic Eqs. for Motion with
Constant Acceleration
y = y0 + vy0 t + ½ ay t 2 (1)
vy = v0y + ay t (2) vy
2 = v0y2 + 2ay (y – y0) (3)
x-t
y-t
Kinematic Eqs. are related by derivatives and integrals.
29
[Note] Eq. (3) can be obtained from (1) and (2) by eliminating t.
Kinematics (1D)
A ball is dropped from the Leaning Tower of Pisa.
It starts from rest at height 50.0 m and falls
freely.
a) Compute its position after 2.00 s.
b) Find the velocity after 2.00 s.
c) Find the time at which the
ball hits the ground.
d) Find the velocity just before
the ball hits the ground.
e) Sketch a y-t graph.
Problem I-1
30
Kinematics (1D)
?
?
?
ISEE
Typical “1-D Motion” Problem
31
Kinematics (1D)
(a) Draw a diagram;
(b) Motion with
constant
acceleration
Kinematic eqs.
Write down kinematic eqs.
Solve the equations.
Typical “1-D Motion” Problem
ISEE
32
Kinematics (1D)
D.A.D. Identify unknowns! ay = –9.80 m/s2
y0 = 50.0 m vy0 = 0.00 m/s
2 unknowns 2 equations if t is given.
0 m
50.0 m
y
y = ? vy = ? @ t = 2.00 s
Problem 1 Solution ISEE
33
Kinematics (1D)
y = y0 + vy0 t + ½ ay t 2 ?
vy = vy0 + ay t ? vy
2 = vy02 + 2ay (y – y0) ?
ay = –9.80 m/s2
y0 = 50.0 m vy0 = 0.00 m/s
ISEE Problem 1 Solution (Cont’d)
34
Kinematics (1D)
(a)Eq. 1
(b)Eq. 2
(c) Eq. 1
(d)Eq. 3
OR
Eq. 2
y = y0 + vy0 t + ½ ay t 2 y = 50.0 + ½ (–9.80) t 2 (1)
vy = vy0 + ay t vy = (–9.80) t (2)
vy2 = vy0
2 + 2ay (y – y0) vy2 = 2 (–9.80) (y – 50.0) (3)
ISEE Problem 1 Solution (Cont’d)
35
Kinematics (1D)
D.A.D. Identify unknowns! ay = 9.80 m/s2
y0 = 0.0 m vy0 = 0.00 m/s
2 unknowns 2 equations if t is given.
50.0 m
0 m
y
y = ? vy = ? @ t = 2.00 s
ISEE Problem 1 Solution (Cont’d)
37
Kinematics (1D)
(a)Eq. 1
(b)Eq. 2
(c) Eq. 1
(d)Eq. 3
OR
Eq. 2
y = y0 + vy0 t + ½ ay t 2 y = 0.0 + ½ (9.80) t 2 (1)
vy = vy0 + ay t vy = (9.80) t (2)
vy2 = vy0
2 + 2ay (y – y0) vy2 = 2 (9.80) (y – 0.0) (3)
ISEE
38
Kinematics (1D)
Problem I-2: A person standing at the edge
of a cliff throws a ball vertically upward
with an initial speed of v0 = 15.0 m/s from
the edge of a cliff that is h = 40.0 m above
the ground. The acceleration due to gravity
is g = 9.80 m/s2 pointing down. Ignore air
friction.
a. (5 pts) How long does it take the ball to reach the ground?
b. (5 pts) What is the speed of the ball just before it strikes the ground?
c. (10 pts) Sketch y-t, vy-t, and ay-t graphs for the motion.
d. (5 pts) If another ball is thrown vertically downward with the same initial speed, the ball to hit the ground with the greater speed is the one initially thrown:
(i) upward.
(ii) downward.
(iii) neither – they both hit at the same speed.
h = 40.0 m
40
Kinematics (1D)
h = 40.0 m
ISEE
Hints for Problem I-2
41
Problem I-3: A person standing at the edge
of a cliff throws a ball vertically upward
with an initial speed of v0 = 15.0 m/s from
the edge of a cliff that is h = 40.0 m above
the ground. The acceleration due to gravity
is g = 9.80 m/s2 pointing down. Ignore air
friction.
Kinematics (1D)
h = 40.0 m
y = 0
y
y = h
a. (5 pts) How long does it take the ball to reach the ground?
b. (5 pts) What is the speed of the ball just before it strikes the ground?
c. (10 pts) Sketch y-t, vy-t, and ay-t graphs for the motion.
d. (5 pts) If another ball is thrown vertically downward with the same initial speed, the ball to hit the ground with the greater speed is the one initially thrown:
(i) upward.
(ii) downward.
(iii) neither – they both hit at the same speed.
43
Kinematics (1D)
You can find the velocity at t = 4.00 s in the same manner as in the previous example.
Problem I-4
44
Kinematics (1D)
Problem I-5: Using vx2 = v0x
2 + 2 ax (x – x0)
Motion with constant acc. (negative : deacceleration)
x – x0
Q: Braking Distance ?
vx
t
v0x
t0
45
x
Kinematics (1D)
Problem I-6
An antelope moving with constant acceleration
covers the distance between two points that are
80.0 m apart in 7.00s. Its speed as it passes the
second point is 15.0 m/s.
0) Have a diagram
a) What is the speed at the first point?
b) What is the acceleration?
46
Kinematics (1D)
A sports car is advertised to be able to stop in
a distance of 55.0 m from a speed of 100 km/h.
a) What is its acceleration in m/s2?
b) How many g’s is this?
1) Conversion: 100 km/h ?? m/s
2) Use vx2 – v0x
2 = 2 ax (x – x0) to find ax.
3) |ax|/g where g = 9.80 m/s2
?
?
?
Problem I-7: How Many g’s?
48
Kinematics (1D)
Problem II-1 (2-body problem): A police car at rest, passed by a
speeder traveling at a constant 100 km/hr, takes off in hot
pursuit. The police officer catches up to the speeder in
800 m, maintaining a constant acceleration.
a) How long did it take the police officer to overtake the speeder?
b) What was the acceleration of the police car?
c) What was the speed of the police car at the overtaking point?
vxs0 = 27.8 m/s
Hint: Kinematic eqs. for each body.
d = 800 m
D.A.D
ISEE
50
Kinematics (1D)
x = x0 + vx0 t + ½ ax t2 d = ½ ap t 2 (1)
vx = vx0 + ax t vxp = ap t (2)
vx2 = vx0
2 + 2ax (x – x0) vxp2 = 2 ap d (3)
Problem II-1 Solution ISEE
51
Kinematics (1D)
x = x0 + vx0 t + ½ ax t2 d = ½ ap t 2 (1)
vx = vx0 + ax t vxp = ap t (2)
vx2 = vx0
2 + 2ax (x – x0) vxp2 = 2 ap d (3)
Problem II-1 Solution
d = vxs0 t (4)
vxs = vxs0 (5)
vxs2 = vxs0
2 (6)
ISEE
52
Kinematics (1D)
Where should the professor be when you release the egg?
Egg Drop
Problem II-2
54
Kinematics (1D)
(a) Draw a diagram
(b) Motion with
constant
acceleration
(c) 2 bodies, so 2 sets of
eqs.
Write down kinematic eqs.
Solve the equations.
Typical “1-D Motion” Problem
ISEE
55
Kinematics (1D)
0 m
46.0 m
1.8 m
DaD at t =0
ay = –9.80 m/s2
y0 = 46.0 m v0y = 0.00 m/s
ax = 0 m/s2
x0 = ? m v0x = 1.20 m/s
y
x d = ?
Where should the professor be when you release the egg?
ISEE
56
Kinematics (1D)
. and find and equations two Solve
unknowns. Two and equations Two :Check
801)0980()046(
0 (1.20)
; Hit :Concept
1.80 ; (1.20) [Prof]
)0980()046( ; 0 [Egg]
0
2
0
profeggprofegg
prof0prof
2
eggegg
tx
.t...
tx
yyxx
ytxx
t...yx
ISEE
ISEE 58
Kinematics (1D)
some
Diagnostic Test
61
Kinematics (1D)
Straight-line Parabola
Describe the motion from the graph: Look at slope Speed Zero? Constant? Increase? Decrease? v-t graph
Diagnostic Test
63
Kinematics (1D)
arbitrary scale
v
t
How do you grade this (0-10 scale)?
Student A’s Graph
64
Kinematics (1D)
Student B’s Graph
How do you grade this (0-10 scale)?
65
Kinematics (1D)
Student C’s Graph
How do you grade this (0-10 scale)?
66
Kinematics (1D)
Problem I-G1
69
Kinematics (1D)
Problem I-G2
71
Kinematics (1D)
MP
72
Kinematics (1D)
How to study Chap. 2 Problems
Key Categiries MP TextBook
I 1 body with g 6(2.37), 11 Examples 2.6, 2.7, 2.8;
2.35, 2.37, 2.44, 2.81, 2.84
1 body with a 3(a=0), 4(2.14), 7,
8(2.51)
Examples 2.1, 2.3, 2.9;
2.14, 2.51
Graphs 1, 2, 5 Examples 2.2;
2.10, 2.11
II 2 bodies with g 2.49, 2.94
2 bodies with 2 a’s 9, 10 Examples 2.5;
2.32(graph), 2.80, 2.82, 2.93
III 1 body with 2 a’s
(Repeat Type I, twice)
2.40, 2.86
73
Kinematics (1D) 74
1
75
2
3
Kinematics (1D) 76
4
5
6
Kinematics (1D) 77
7
8
9
Kinematics (1D) 78
10
11