PHY10 Lesson 2Motion Along a Straight Line

Position, Displacement, and Average VelocityConsider the motion of a car below.

At time t1 = 1s , the car is at position x1 =19m and At time t2 = 4s , the car is at position x2 = 277m.The displacement of the car is:

mmmxxx 2581927712

The time interval is: sssttt 31412

The average velocity of the car is: t

xvav

12

12

tt

xxvav

sms

mvav /86

3

258

Graphs of Motion Part 1: Position vs. Time

The object is at rest.

x (

m)

t (s)

x (

m)

t (s)

x (

m)

t (s)x (

m)

t (s)

x (

m)

t (s)

The object is moving with constant velocity in the positive direction.

The object is moving with constant velocity in the negative direction.

The object is moving with increasing velocity (accelerating).

The object is moving with decreasing velocity (decelerating).

EXAMPLE 1:

smss

mmmv

smm

smmave /4.4

7040

480)280()200(

/4280

/5200

(a) average speed

(b) average velocity

smss

mmmv

smm

smmave /73.0

7040

80)280()200(

/4280

/5200

Instantaneous Velocity

instantaneous velocity (v) is the velocity of a body at a specific instant of time or at any specific point along the path.

instantaneous velocity (v) is the limit of average velocity of a body as the time interval approaches zero.

t

xtv

t

0lim)(

instantaneous velocity (v) is the derivative of displacement with respect to time.

dt

dxtv )(

EXAMPLE 2:

3322 )/120.0()/40.2()( tsmtsmtx (a)

0)0)(120.0()0)(40.2()0( 32 x

mx 120)10)(120.0()10)(40.2()10( 32

smm

m

tt

xxvave /12

010

0120

12

12

23232 )/360.0()/80.4()120.040.2()( tsmtsmttdt

d

dt

dxtv (b)

0)0)(360.0()0)(80.4()0( 2 v

smv /15)5)(360.0()5)(80.4()5( 2

smv /12)10)(360.0()10)(80.4()10( 2

(c) 2360.080.40 tt st 3.13

EXAMPLE 2:

Average and Instantaneous Acceleration

acceleration (a) is the time rate of change in velocity.

average acceleration (aav) 12

12

tt

vv

t

vaav

instantaneous acceleration (a) is the limit of average acceleration as the time interval approaches zero.

dt

dvta )(

t

vta

t

0lim)(

23)/100.0()/00.3()( tsmsmtv (a)

EXAMPLE 3:

smsmsmv /00.3)0)(/100.0()/00.3()0( 23

smssmsmv /500.0)5)(/100.0()/00.3()5( 23 2

12

12 /500.0)05(

)/500.0()/00.3(sm

s

smsm

tt

vvaave

tsmta )/200.0()( 3(b) 0)0)(/200.0()0( 3 sma

23 /00.1)5)(/200.0()5( smssma

(c)

EXAMPLE 3:

Graphs of Motion Part 2: Velocity vs. Time

The object is at rest.

v (

m/s

)

t (s)

v (

m/s

)

t (s)v (

m/s

)

t (s)

v

(m/s

)

t (s)

The object is moving with constant velocity in the positive direction.

The object is moving with increasing velocity (uniformly accelerating).

The object is moving with decreasing velocity (uniformly decelerating).

v

(m/s

)

The object is moving with constant velocity in the negative direction.

t (s)

Graphs of Motion Part 3: Acceleration vs. Time

a

(m/s

2)

t (s)

The object is moving with constant velocity in the positive direction.

The object is moving with constant velocity in the negative direction.

a

(m/s

2)

t (s)

The object is moving with increasing velocity (uniformly accelerating).

a

(m/s

2)

t (s)

The object is moving with decreasing velocity (uniformly decelerating).

a

(m/s

2)

t (s)

Uniformly Accelerated Motion (UAM)

uniformly accelerated motion is motion with constant acceleration.

t

xvav

Equations of Uniformly Accelerated Motion

2

vvv oav

t

vva o

Equation 1:

Equation 2:

Equation 3:

assuming that t0 = 0.

assuming that t0 = 0.

dt

dxv

Equations of Uniformly Accelerated Motion

Recall:

Equation 4:

Equation 5:

dtvdx t

t o

t

t

x

x ooo

dtatvvdtdx )(

221 attvxx oo 2

21 attvx o

dt

dva Recall: dtadv adxdta

dt

dxvdv

v

v

x

x oo

vdva

dx1

22

1 22o

o

vv

axx

a

vvx o

2

22

(a)

EXAMPLE 4:

GIVEN: x = 70.0m ; t = 7.00 s ; v = 15.0 m/sFIND: (a) vo and (b) a

sms

m

t

xvav /0.10

7

70

sm

vsmvvv ooav /0.10

2

/15

2

smsmsmvo /00.5)/15()/10(2

(b) 2/43.17

)/5()/15(sm

s

smsm

t

vva o

#5:

#5:

Chapter 02, Problem 015A particle's position is given by x = 10.0 - 12.00t + 3t2, in which x is in meters and t is in seconds. (a) What is its velocity at t= 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer "0". (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer "0".

Chapter 02, Problem 017The position of a particle moving along the x axis is given in centimeters by x = 9.17 + 1.17 t3, where t is in seconds. Calculate(a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00 s; (c) the instantaneous velocity at  t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s.

Chapter 02, Problem 038(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 23.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?

Chapter 02, Problem 040You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of v0 = 55 km/h; your best deceleration rate has the magnitude a = 5.18 m/s2. Your best reaction time to begin braking is  t = 0.75 s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 km/h if the distance to the intersection and the duration of the yellow light are (a) 37 m and 2.9 s, and (b) 32 m and 1.6 s? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).

Free Fall (UAM along the y-axis)

free fall is motion under the action of the force of gravity alone (air resistance is neglected).

• a freely-falling body has a constant acceleration called the acceleration due to gravity g = - 9.80 m/s2 (always directed downward).

t

yvav

Equations of Free Fall

2

vvv oav

t

vvg o

Equation 1:

Equation 2:

Equation 3:

221 gttvy o

g

vvy o

2

22

Equation 4:

Equation 5:

NOTE: Follow correct sign convention. All quantities with downward direction should have a negative sign.

downward velocity: - vupward velocity: +vdownward displacement: - yupward displacement: + yacceleration: g= -9.80 m/s2

#6:

#7:

ANSWERS:

(a) y1 = 10.1m; v1 = 5.2 m/s and y4 = -18.4 m; v1 =-24.2 m/s

(b) v =±11.3 m/s (c) y =+11.5 m (d) a = g = - 9.8 m/s2

#8

Chapter 02, Problem 054A stone is dropped into a river from a bridge 45.4 m above the water. Another stone is thrown vertically down 1.91 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Answer: 34.5 m/s

Chapter 02, Problem 060A rock is thrown vertically upward from ground level at time t = 0. At t = 1.7 s it passes the top of a tall tower, and 1.2 s later it reaches its maximum height. What is the height of the tower?

Answers: 34 m

Chapter 02, Problem 062A basketball player grabbing a rebound jumps 77 cm vertically. How much total time (ascent and descent) does the player spend(a) in the top 10 cm of this jump and (b) in the bottom 10 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

Answers: (a) 0.286 s (b) 0.0533s

Chapter 02, Problem 063A drowsy cat is looking at a window from across the room, and sees a flowerpot that sail first up and then down past the window. The pot is in view for a total of 0.46 s, and the top-to-bottom height of the window is 2.02 m. How high above the window top does the flowerpot go?

Answers: 2.99 m

Projectile MotionA body moving with constant horizontal velocity while in free fall

is called a projectile and its motion is called projectile motion.

velocity of projectile:

22yx vvv

cosox vv

gtvv oy sin

x

y

v

vtan

Motion along x-axis: constant velocity

t

xvx

||

2sin2

g

vR o

Motion along y-axis: free fall

gtvv oy sin

221sin gttvy o

g

vvy yoy

2

22

||2

sin 2

max g

vyH o

Chapter 04, Problem 023A projectile is fired horizontally from a gun that is 35.0 m above flat ground, emerging from the gun with a speed of 170 m/s. (a)How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c)What is the magnitude of the vertical component of its velocity as it strikes the ground?

Answers: (a) 2.67 s (b) 454 m (c) 26.2 m/s

Answers: (a) 9.96 s (b) 837 m

Chapter 04, Problem 027A certain airplane has a speed of 261.5 km/h and is diving at an angle of θ = 29.0° below the horizontal when the pilot releases a radar decoy (see the figure). The horizontal distance between the release point and the point where the decoy strikes the ground is d = 633 m. (a) How long is the decoy in the air? (b) How high was the release point?

Chapter 04, Problem 028In the figure, a stone is projected at a cliff of height h with an initial speed of 49.0 m/s directed at an angle θ0 = 55.0° above the horizontal. The stone strikes at A, 5.76 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.

Chapter 04, Problem 045In the figure, a ball is launched with a velocity of magnitude 8.00 m/s, at an angle of 47.0° to the horizontal. The launch point is at the base of a ramp of horizontal lengthd1 = 6.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp. (a) Does the ball land on the ramp or the plateau? When it lands, what are the (b)magnitude and (c) angle of its displacement from the launch point?

Answers: (a) 17.8 m/s (b) in the river, 28.5 m from the near bank