Chapter 2: Motion in One Dimension
Ch2.1 Position, Velocity, and Speed
A. Position, Distance, Displacement
1. Position along a number line
0
+
2. Distance is total length traveled
3. Displacement x = xf - xi
B. Average Speed
distance traveled / elapsed time
C. Average velocity = vx = x/ t
This question is asking about displacement. What is
the change in the distance x?
Common misconceptions - Distance traveled is
determined by giving the final position or the
student incorrectly identifies the initial position as
zero.
Correct answer
Concept Question 2.1
You travel from your house to the grocery store,
shop, bring the groceries to your friend’s house and
return home. The distance you travel is?
A.
B.
C.
D.
E.
0 mi
6.4 mi
8.5 mi
10.7 mi
12.8 mi
Concept Question 2.2
You travel from your house to the grocery store,
shop, bring the groceries to your friend’s house and
return home. Your displacement is?
A.
B.
C.
D.
E.
0 mi
6.4 mi
8.5 mi
10.7 mi
12.8 mi
Chapter 2: Motion in One Dimension
Ch2.1 Position, Velocity, and Speed
Concept Question 2.3
Do QQ2.1 p. 22 with clickers
Ways to represent data
Table
t(s)
x = t^2
Graph
x(m)
0
1
2
3
4
5
0
1
4
9
16
25
Position vs. Time
30
25
20
x(m)
Equation
15
Series1
10
5
0
0
1
2
3
t(s)
4
5
6
Equation
Table
t(s)
x(m)
0
1
2
3
4
5
0
1
4
9
16
25
Position vs. Time
30
25
20
x(m)
x = t^2
Graph
15
Series1
10
5
0
0
1
2
3
4
5
6
t(s)
Vx (from t = 2 to t = 4) =
(16m – 4m) / (4s – 2s) = 6 m/s
Slope = rise / run = 12 m / 2 s = 6 m/s
Concept Question 2.4
The average velocity between t = 1 and t = 2 is?
A. -1 m/s
E. 2 m/s
B. 0.5 m/s
C. 1 m/s
D. -0.5 m/s
Chapter 2: Motion in One Dimension
Ch2.2 Instantaneous Velocity
A. Instantaneous velocity definition
vx
=
lim x/t
t 0
B. Graphical Interpretation
Vx is the slope of the line tangent to
the x vs. t curve at the instant of time
in question
Look at xt2inst.xls
P2.5 (p. 46)
Concept Question 2.5
A.
B.
C.
D.
Chapter 2: Motion in One Dimension
Ch2.4 Acceleration
A. Average acceleration
ax = v/t
B. Instantaneous Acceleration
ax
=
lim v/t
t 0
ax is the slope of the line tangent to
the vx vs. t curve at the instant of time
in question
Concept Question 2.6
The average acceleration between t = 0 and t = 20
is?
A. 1 m/s2
B. -1 m/s2
C. 0.5 m/s2
D. 2 m/s2
E. 0.2 m/s2
P2.16 (p. 47)
Fig. P2.17, p.51
Chapter 2: Motion in One Dimension
Ch2.5 Motion Diagrams
Moving Man Simulation
Concept Question 2.7
A
B
C
D
E
F
Acceleration occurs when there is a change in
velocity. The velocity is constant 1 m/s between 0
and 2 seconds and it is zero from 2 to 4 seconds.
The only change is at t = 2 seconds.
Common misconception - Student interprets
sloping up (or down) on a position graph to mean
the object is speeding up (or slowing down).
Correct answer
Answer [b:b] is correct.
Many picked answer [c:d], [c:c], or [c:b].
What is wrong with those choices?
Answer b is correct.
Some picked answer c.
Equations for Constant
Acceleration Only
1. vxf = vxi + axt
2. xf = xi + (vxi + vxf) t / 2
3. xf = xi + vxi t + axt2/2
4. vxf2 = vxi2 + 2ax(xf – xi)
Assuming the conditions: ti = 0, tf = t,
x(0) = xi and v(0) = vxi and ax is constant.
Strategy
1. Convert units if necessary.
2. Draw a picture that includes given information.
3. Pick coordinate origin.
4. List givens and finds.
5. Select equation based on results of 3 and
solve.
6. Check if answer reasonable.
CT2.8: You are throwing a ball straight up in
the air. At the highest point, the ball’s
A. velocity and acceleration are
zero.
B. velocity is nonzero but its
acceleration is zero.
C. acceleration is nonzero, but its
velocity is zero.
D. velocity and acceleration are
both nonzero.
The ball is continuing to lose velocity during the
whole trip. When it turns around at the top, the
velocity is momentarily zero, but the velocity is still
decreasing as it becomes greater negative.
Common misconception - Student concludes that
if an object has zero speed, even for an instant, it
also has zero acceleration. (This instant may
appear at the starting point, ending point, or a turn
around point.)
Correct answer
The symmetry of throwing a ball upward.
0 m/s
t=2s
19.6 m/s
t=0s
19.6 m/s
t=4s
A woman is reported to have fallen 144 ft from
the 17th floor of a building, landing on a metal
ventilator box, which she crushed to a depth of 18
in. She suffered only minor injuries. Neglecting
air resistance, calculate (a) the speed of the
woman just before she collided with the
ventilator, (b) her average acceleration while in
contact with the box, and (c) the time it took to
crush the box.
Ct2.9: If you drop an object in the absence
of air resistance, it accelerates downward at
9.8 m/s2. If instead you throw it downward,
its downward acceleration after release is
A. less than 9.8 m/s2.
B. 9.8 m/s2.
C. more than 9.8 m/s2.
Ct2.10: A person standing at the edge of
a cliff throws one ball straight up and
another ball straight down at the same
initial speed. Neglecting air resistance,
the ball to hit the ground below the cliff
with the greater speed is the one
initially thrown
A. upward.
B. downward.
C. neither—they both hit at the
same speed.
x = (vx)(t) =
area of triangle
x =
(vx)(t) =
area of
trapezoid
x = (vx)(t) =
area of trapezoid
Statements Connecting Motion
Variables that are Always True
• The instantaneous velocity is the slope of
the tangent to the x vs. t curve.
• The instantaneous acceleration is the
slope of the tangent to the v vs. t curve.
• The change in position is the area under
the v vs. t curve.
• The change in velocity is the area under
the a vs. t curve.