1. Position along a number line

- 0 +

2. Distance is total length traveled

3. Displacement x = xf - xi

Ch2-1 Position, Distance, and Displacement

Chapter 2: One-Dimensional Kinematics

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

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

A

B

C

D

C2.3

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

A

B

C

D

C2.4

Common misconceptions - Student incorrectly identifies the initial position as zero or determines distance by adding two or more positions during the motion. Correct answer

A. Average Speed

distance traveled / elapsed time

B. Average velocity

vav = displacement / elapsed time

vav = (xf–xi)/(tf–ti) = x/t

Ch2-1 Average Speed and Velocity

Chapter 2: One-Dimensional Kinematics

A

B

C

D

C2.5

Common misconceptions - Student interprets an upward (or downward) sloping graph to mean the object is going up hill (or downhill) or does not distinguish position and/or distance from speed. Correct answer

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

Vav (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 2.6

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

Do P2.13 (p. 48)

Chapter 2: One-Dimensional Kinematics

A B C

D E F

Concept Question 2.7

Ch2-3 Instantaneous Velocity

A. Instantaneous velocity definition

v = lim x/t t 0

B. Graphical Interpretation

v is the slope of the line tangent to the x vs. t curve at the instant of time in question

Chapter 2: One-Dimensional Kinematics

Look at xt2inst.xls

Chapter 2: One-Dimensional Kinematics

A.

B.

C.

D.

Concept Question 2.8

Look at P2-27&26.xls

Chapter 2: One-Dimensional Kinematics

Ch2-4 Acceleration

A. Average acceleration

aav = (vf-vi)/(tf-ti) = v/tB. Instantaneous Acceleration

a = lim v/t t 0

a is the slope of the line tangent to thev vs. t curve at the instant of time in question

Chapter 2: One-Dimensional Kinematics

Concept Question 2.9

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

Do P2.29 (p. 49)

Chapter 2: One-Dimensional Kinematics

Ch2-4 Acceleration

Do P2.33(p. 49)

Area = (vav)(t)

Area = (vav)(t)

trapezoid

Area = (vav)(t)

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.

Concept Question 2.10P2.35a: Assuming constant deceleration, if you double your driving speed from 16 m/s to 32 m/s, the time required to come to a stop increases by a factor of

A. two.B. four.

Do 2.35b (p. 49)

Chapter 2: One-Dimensional Kinematics

2-5 Motion with Constant Acceleration

Assuming the initial conditions at t = 0, x = x0 and v = v0 and a is constant.

Equations for Constant Acceleration Only

1. v = v0 + at

2. vav = (v0 + v) / 2

3. x = x0 + (v0 + v) t / 2

4. x = x0 + v0 t + at2/2

5. v2 = v02 + 2a(x – x0)

Assuming the initial conditions at t = 0, x = x0 and v = v0 and a is constant.

A.

B.

C.

CT 2.11

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) or Student concludes that if an object has a speed, it must be accelerating. Correct answer

Almost everyone picked c for the first question.

Many picked answer [c:c] or [c:d]

What is wrong with those choices?

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

1. v = v0 + at

2. vav = (v0 + v) / 2

3. x = x0 + (v0 + v) t / 2

4. x = x0 + v0 t + at2/2

5. v2 = v02 + 2a(x – x0)

2-6 Applications of the Equations of MotionDo 2.42 (p. 50)

Similar to 2.43

Do 2.53 (p. 50)

Do 2.54 (p. 50)

Concept Question 2.12: P2.53b:After braking for half the time found in part a, your speed is

A. 8 m/s.B. > 8 m/s.C. < 8 m/s.

Concept Question 2.13: P2.54b:In the first 0.05 s, does the tongue extend

A. 8 cm?B. > 8 cm?C. < 8 cm?

1. v = v0 + at

2. vav = (v0 + v) / 2

3. y = y0 + (v0 + v) t / 2

4. y = y0 + v0 t + at2/2

5. v2 = v02 + 2a(y – y0)

Do 2.68 (p. 52)

similar to 2.78 and 2.97d

Do 2.75 (p. 52)

similar to 2.70

CT2.14: 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.

2-7 Freely Falling Objects

19.6 m/st = 4 s

19.6 m/st = 0 s

The symmetry of throwing a ball upward.

0 m/st = 2 s

Ct2.15: 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.

Discuss P2.97 (p.53).What is v0?

Ct2.13: I have started the WebAssign homework for Chapter 2.

A. YesB. No