A. Position, Distance, Displacement 1. Position along a number line - 0 + 2. Distance is total length traveled 3. Displacement x = x f - x i B. Average Speed distance traveled / elapsed time C. Average velocity = v = x/ t Ch2.1 Position, Velocity, and Speed Chapter 2: Motion in One Dimension
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A. Position, Distance, Displacement 1. Position along a number line - 0 + 2. Distance is total length traveled 3. Displacement x = x f - x i B. Average.
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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
Ch2.1 Position, Velocity, and Speed
Chapter 2: Motion in One Dimension
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. 0 miB. 6.4 miC. 8.5 miD. 10.7 miE. 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. 0 miB. 6.4 miC. 8.5 miD. 10.7 miE. 12.8 mi
Ch2.1 Position, Velocity, and Speed
Concept Question 2.3
Do QQ2.1 p. 22 with clickers
Chapter 2: Motion in One Dimension
t(s) x(m)x = t^2 0 0
1 12 43 94 165 25
Position vs. Time
0
5
10
15
20
25
30
0 1 2 3 4 5 6
t(s)
x(m
)Series1
Ways to represent data
Equation Table Graph
t(s) x(m)x = t^2 0 0
1 12 43 94 165 25
Position vs. Time
0
5
10
15
20
25
30
0 1 2 3 4 5 6
t(s)
x(m
)
Series1
Vx (from t = 2 to t = 4) =
(16m – 4m) / (4s – 2s) = 6 m/s
Slope = rise / run = 12 m / 2 s = 6 m/s
Equation Table Graph
Concept Question 2.4
The average velocity between t = 1 and t = 2 is?
A. -1 m/s B. 0.5 m/s C. 1 m/s D. -0.5 m/s
E. 2 m/s
Ch2.2 Instantaneous Velocity
Chapter 2: Motion in One Dimension
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)
A.
B.
C.
D.
Concept Question 2.5
Ch2.4 Acceleration
Chapter 2: Motion in One Dimension
A. Average acceleration
ax = v/tB. 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
Fig. P2.17, p.51
P2.16 (p. 47)
Ch2.5 Motion Diagrams
Moving Man Simulation
Chapter 2: Motion in One Dimension
A B C
D E F
Concept Question 2.7
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 = vxi
2 + 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
19.6 m/st = 4 s
19.6 m/st = 0 s
The symmetry of throwing a ball upward.
0 m/st = 2 s
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 at9.8 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 MotionVariables 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.