John Lin09.04.13

Pd. 4A

Chapter 2: Motion in One Dimension

Introduction

● kinematics: study of motion● types of motion:

translation: car traveling on highway rotation: Earth’s spin on its axis vibrational: pendulum

● particle model: describe moving object as a particle particle: point-like object (has mass, infinitesimal size)

Chapter 2.1: Position, Velocity, & Speed

● position x: location of particle w/ respect to reference point (origin of coordinate system)● displacement ∆x: ∆ x ≡ x f −x i

● distance: length of path followed by a particle

● average velocity: vx ,avg ≡∆ x∆ t

, displacement over time

● average speed: vavg≡d

∆ t, distance over time, no direction

● vector quantity: direction & magnitude displacement, velocity, acceleration

● scalar quantity: has magnitude but no direction

Chapter 2.2: Instantaneous Velocity & Speed

● instantaneous velocity :v x ≡ lim ¿∆ t → 0∆ x∆ t

=dxdt

¿

slope of tangent line to x-t graph velocity: instantaneous velocity

● instantaneous speed: magnitude of instantaneous velocity

Chapter 2.7: Freely Falling Objects

● if air resistance is absent (free fall), all objects dropped near Earth’s surface fall w/ same constant acceleration (gravity)

● freely falling object: moving freely under influence of gravity alone● free-fall acceleration g: decreases w/ increasing altitude, changes w/ latitude

g=9.80 m/s2

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John Lin09.04.13

Pd. 4A

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John Lin09.04.13

Pd. 4A

Kinematic Equations for Motion of a Particle Under Constant Acceleration

1. vxf =v xi+ax t

2. x f =x i+12(v xi+vxf ) t

3. x f =x i+vxi t +12

a t 2

4. vxf2=v xi

2+2 ax(x f−x i)

Problems

5. A position-time graph for a particle moving along the x-axis is shown in Figure P2.5. a. Find the average velocity in the time interval t = 1.50 s to t = 4.00 s.

vx ,avg=x f −x i

∆ t=2−8

2.5=−2.4 m/s

b. Determine the instantaneous velocity at t = 2.00 s by measuring the slope of the tangent line shown in the graph.

vx=m= 0−133.5−0

=−3.7 m/s

c. At what value t is the velocity zero.t=4.0s

9. The North American and European plates of the Earth’s crust are drifting apart with a relative speed of about 25 mm/yr. Take the speed as constant and find when the rift between them started to open, to reach a current width of 2.9 × 103 mi.

2.9 ×103 mi . ×1609344 mm

1mi .=4 667 097 600 mm

t= Drate

=4 667 097 600 mm25 mm / yr

=186 683 904 yr=1.9 ×108 years ago

22. A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. a. Find its original speed.

vavg=v i+v f

2=∆ x

∆ t= 40 m

8.50 s=4.7 m /s

v i+2.8

2=4.7 m /s

v i+2.8=9.4 m /sv i=6.6 m /s

b. Find its acceleration.

a=∆ v∆ t

=2.8 m /s−6.6 m / s8.50 s

=−0.45 m / s2

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John Lin09.04.13

Pd. 4A

30. In the particle under constant acceleration model, we identify the variables and parameters vxi, vxf, ax, t, and xf – xi. Of the equations in Table 2.2, the first does not involve xf – xi, the second does not contain ax, the third omits vxf, and the last leaves out t. So, to complete the set, there should be an equation not involving vxi.a. Derive it from the others.

vxf =v xi+ax t

vxi=v xf−ax t

x f =x i+vxi t +12

ax t 2=x i+ (v xf −ax t ) t+ 12

ax t 2=x i+v xf t−ax t2+ 12

ax t 2

x f =x i+vxf t−12

ax t 2

b. Use the equation to solve Problem 29 in one step. Problem #29: The driver of a car slams on the brakes when he sees a tree blocking the road. That car slows uniformly with an acceleration of –5.60 m/s2 for 4.20 s, making straight skin marks 62.4 m long, all the way to the tree. With what speed does the car then strike the tree?

x f =x i+vxf t−12

ax t 2

62.4 m=0 m+v xf ×4.20 s−12

×−5.60 m /s2×(4.20 s)2

62.4 m=v xf × 4.20 s+49.392 m

vxf =62.4 m−49.392m

4.20 s=3.10 m /s

31. A glider of length l moves through a stationary photogate on an air track. A photogate is a device that measures the time interval ∆ t d during which the glider blocks a beam of infrared

light passing across the photogate. The ratio vd=l

∆ t d is the average velocity of the glider

over this part of its motion. Suppose the glider moves with constant acceleration.a. Argue for or against the idea that vd is equal to the instantaneous velocity of the glider

when it is halfway through the photogate in space.Unless the acceleration is 0, vd is not equal to the instantaneous velocity of the

glider when it is halfway through the photogate in space. Since the glider moves with constant acceleration, velocity changes steadily with time, not space. Since velocity changes, the time it takes to travel the first half of the photogate in space differs from that to travel the second half.

b. Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time.

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John Lin09.04.13

Pd. 4A

Unless the acceleration is 0, vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time. Since the glider moves with constant acceleration, velocity changes steadily with time.

Let vx be the instantaneous velocity of the glider when it is halfway through the photogate in time. If acceleration is constant, then: vx=v i+at , v f=v i+2 at

vd=v i+v f

2=

v i+v i+2 at

2=v i+at=v x

Therefore, vd is the instantaneous velocity when the glider is halfway through the photogate in time.

43. A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The second student catches the keys 1.50 s later. a. With what initial velocity were the keys thrown?

x f −x i=v xi t+12

a t 2

4.00 m=v xi× 1.50 s+12

×−9.80 m /s2×(1.50 s)2

vxi ×1.50 s=15.025 m

vxi=10.0 m/s up

b. What was the velocity of the keys just before they were caught?

vxf =v xi+at=10.0 m / s−9.80 m /s2×1.50 s=−4.69 m / s=4.69 m /s down

59. At t=0, one athlete in a race running on a long, straight track with a constant speed v1 is a

distance d1 behind a second athlete running with a constant speed v2.a. Under what circumstances is the first athlete able to overtake the second athlete?

v1>v2, d1 must be great enough that the first runner have time to catch up to the second athlete.

b. Find the time t at which the first athlete overtakes the second athlete, in terms of d1 , v1 ,∧v2.d x: distance first runner travels, d y: distance second runner travelsd x=v1 t , d y=v2t

d x=d y+d1

d1=v1t−d y=v1 t−v2 t

t=d1

v1−v2

c. At what minimum distance d2 from the leading athlete must the finish line be located so

that the trailing athlete can at least tie for first place? Express d2 in terms of d1 , v1 ,∧v2 .

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John Lin09.04.13

Pd. 4A

d y=v2 t=v2×d1

v1−v2

=v2 d1

v1−v2

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