Chapter 3 Describing Motion

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Chapter 3 Describing Motion. Scalar A quantity that has only magnitude. Time (t) Mass (m) Distance (d) Temperature (T) Speed (s). Chapter 3 Describing Motion. Vector A quantity with magnitude and direction Velocity ( V ) Displacement ( D ) Force ( N ) Weight ( N ). - PowerPoint PPT Presentation

Transcript of Chapter 3 Describing Motion

Chapter 3Describing Motion

Scalar A quantity that has only magnitude.

Time (t) Mass (m) Distance (d) Temperature (T) Speed (s)

Chapter 3 Describing Motion

Vector A quantity with magnitude and direction

Velocity (V) Displacement (D) Force (N) Weight (N)

Chapter 3Describing Motion

Displacement The distance and direction between two positions.

Average Velocity The ratio of the change in position to the time interval during which the change occurred.

Average Speed The ratio of the total distance traveled to the total time

interval.

Chapter 3Describing Motion Average Velocity

Displacement

v dt

d1 do

t1 to

d v t

d

do d1

d

do d1

v

m/s3.80s6s0m50mV

Chapter 3Describing Motion Instantaneous velocity The speed

and direction of an object at a particular time.

The symbol for instantaneous velocity is represented by the symbol v.

Relative velocity The velocity of an object in relationship to another object

Chapter 3Describing Motion Average Acceleration

a vt

v1 vo

t1 to

Solution:The two bicycleswill meet in onehour and if the bee travels at 30 km/hr. The beewill travel 30 km.

When the Smiths went on their vacation they traveled at 40 km/hr. They returned home at a rate of 60 km/hr.

QUESTION:What was their average rate of speed?

40 km/hr>

60 km/hr

Chapter 3 Describing Motion

vdt

d1t1

d2t2

butd1 d2 d and t1 d

40, t2 d

60

v2d

d40 d

60multiplytop andbottom by120

v 2dd

40 d

60120 240d

3d 2d 240d5d 48mph

Solution

a vf vi

t f t ibut if ti = 0, then vf = vi + at

d v t but v vi v f

2so d

v i v f

2

t

d vi v f

2

t

Substitute for vf d vi vi at

2

t

d v it 12

at2

Chapter 3 Describing Motion

vf = vi + at d vi v f

2

t

Solve both equations for t

v f v i

a

2dvi vf

t 2d

vi v f

Set equal to each other

t vf vi

aMultiply across

(vf - vi)(vi + vf) = 2ad

Simplify

vf 2 = vi

2 + 2ad

Chapter 3 Describing Motion

vf = vi + at

d vi v f

2

t

vf 2 = vi

2 + 2ad

Chapter 3 Describing Motion

d v it 12

at2

ag = -9.8 m/s2

ag = -32 ft/s2

tvva if

d v t

Ima Hurryin approaches a stoplight in her car which is moving with a velocity of +30.0 m/s. The light turns yellow, Ima applies the brakesand skids to a stop. If Ima's acceleration is –8.00 m/s2, determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a positive (+) and a negative (–) sign, respectively.)

Chapter 3Describing Motion

Diagram:

Given:vi = +30.0 m/svf = 0 m/sa = –8.00 m/s2

Find:d = ??

Chapter 3Describing Motion

•(0 m/s)2 = (30.0 m/s)2 + 2*(-8.00 m/s2)*d •0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d•(16.0 m/s2)*d = 900 m2/s2 - 0 m2/s2

•(16.0 m/s2)*d = 900 m2/s2

•d = (900 m2/s2)/ (16.0 m/s2)•d = 56.3 m

advv 22i

2f

Chapter 3Describing Motion

Ben Rushin is waiting at a stoplight in his car,When the light turns green, Ben accelerates from rest at a rate of a 6.00 m/s2 for an interval of 4.10 seconds. Determine the displacement of Ben's car during this timeperiod. Given: Find:

vi = 0 m/s d =? t = 4.10 sa = 6.00 m/s2

Chapter 3Describing Motion

•d = (0 m/s)*(4.1 s) + 0.5*(6.00 m/s2)*(4.10 s)2 •d = (0 m) + 0.5*(6.00 m/s2)*(16.81 s2)•d = 0 m + 50.43 m•d = 50.4 m

Chapter 3Describing Motion

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Luke Autbeloe drops a pile of roof shingles from thetop of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

Given: Find:vi = 0.0 m/s t = ?? d = –8.52 ma = –9.8 m/s2

Chapter 3Describing Motion

•-8.52 m = (0 m/s)*(t) + 0.5*(-9.8 m/s2)*(t)2 •-8.52 m = (0 m) *(t) + (-4.9 m/s2)*(t)2

•-8.52 m = (-4.9 m/s2)*(t)2

•(-8.52 m)/(-4.9 m/s2) = t2

•1.739 s2 = t2

•t = 1.32 s

Chapter 3Describing Motion

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Chapter 3 Describing Motion

Motion Applet

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Given: Find:

vi = 26.2 m/s d = ?? vf = 0 m/sa = –9.8 m/s2

Chapter 3Describing Motion

•(0 m/s)2 = (26.2 m/s)2 + 2*(-9.8m/s2)*d •0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2)*d•(-19.6 m/s2)*d = 0 m2/s2 -686.44 m2/s2

•(-19.6 m/s2)*d = -686.44 m2/s2

•d = (-686.44 m2/s2)/ (-19.6 m/s2)•d = 35.0 m

Chapter 3Describing Motion

advv 22i

2f

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until it finally lifts off the ground. Determine the distance traveled before take-off.•Given:•a = 3.20 m/s2

•t = 32.8 s•vi = 0 m/s

•Find:•d = ??

Chapter 3Describing Motion

•d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

•d = 1720 m

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Upton Chuck is riding the Giant Drop at Great America. If Upton free-falls for 2.6 seconds, what will be his final velocity and how far will he fall?

•Find:•d = ?? •vf = ??•d = vi*t + 0.5*a*t2 •d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(2.6 s)2

•d = 33 m

•vf = vi + a*t•vf = 0 + (-9.8 m/s2)*(2.6 s)

•vf = -25.5 m/s (the negative sign indicates direction)

•Given:•a = –9.8 m/s2

•t = 2.6 s•vi = 0 m/s

Chapter 3Describing Motion

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

•vf2 = vi

2 + 2*a*d•(7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)•50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a•(50.4 m2/s2)/(70.8 m) = a•a = 0.712 m/s2

Chapter 3Describing Motion

A car is traveling at 65 km/hr. The brakesare applied and the car stops in 3 seconds.a. What is the car’s acceleration?b. How far does it travel?

vi = 65 km/hrt = 3 svf = 0

vi = 65 km/hr = 18m/s

Chapter 3Describing Motion

Chapter 3Describing Motion

A car is traveling at 65 km/hr. The brakesare applied and the car stops in 3 seconds.a. What is the car’s acceleration?b. How far does it travel?

vi = 18 m/st = 3 svf = 0

ssm

tvv

a if

3/180

2/6 sma

Chapter 3Describing Motion

A car is traveling at 65 km/hr. The brakesare applied and the car stops in 3 seconds.a. What is the car’s acceleration?b. How far does it travel?

2

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22 )3)(/6(21)3)(/18( ssmssmd =27 m

Chapter 3Describing Motion

An airplane must reach a speed of 55 m/sbefore take off. It can accelerate at 12 m/s2.a. How long does the runway have to be?b. How long will this take?

vi = 0a = 12 m/s2

vf = 55 m/s

advv if 222

mavv

d if 1262

22

Chapter 3Describing Motion

An airplane much reach a speed of 55 m/sbefore take off. It can accelerate at 12 m/s2.a. How long does the runway have to be?b. How long will this take?

vi = 0a = 12 m/s2

vf = 55 m/savv if t

ssm

smsm 58.4/12

/0/552

t

Chapter 3Describing Motion

Joe is driving a car at 20 m/s and notices a cow in front of him at a distance of 80 m away. He brakes at a rate of 3 m/s2. What happens to the cow?

vi = 20 m/sa = -3 m/s2

vf = 0

Chapter 3Describing Motion

avv

d if

2

22

)/3(2)/20(0

2

2

smsmd

md 7.66

The cowsurvives!!!!

ag = -9.8 m/s2

For all free falling bodies on earth, theyacceleratedownward at 9.8 m/s2.

Chapter 3Describing Motion

A ball is thrown upward at 49 m/s.a. How long is it in the

air?b. How high does it

go?c. How fast is it going

when it lands?

Chapter 3Describing Motion

How long is it in the air? vi = 49 m/sa = -9.8 m/s2

vf = 0avv if t 2/8.9

/490smsm

t

sec5t Total time in air is 10 sec

Chapter 3Describing Motion

How high does it go?

vi = 49 m/sa = -9.8 m/s2

t = 5 sec

2

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22 )5)(/8.9(21)5)(/49( ssmssmd

md 5.122

Chapter 3Describing Motion

How fast is it going when it lands?

vi = 49 m/sa = -9.8 m/s2

t = 10 secatvv if

)10)(/8.9(/49 2 ssmsmv f

smv f /49

Chapter 3Describing Motion

Chapter 3Describing Motion

Free Fall

A skier travels from A to B to C to D.What is her average velocity and speed?

Chapter 3Describing Motion

Josh travels from A to B to C to D.What is his average velocity and speed?

Chapter 3Describing Motion

min3m140

v =46.7 m/min

min3m420

s =140 m/min

When a traffic light turns green, a waiting car starts off with a constant acceleration of 6 m/s2. At the instant the car begins to accelerate, a truck with constant velocity of 21 m/s passes in the next lane.

a. How far will the car travel before it overtakes the truck? b. How fast will the car be traveling when it overtakes the truck? c. Construct a distance vs time graph for the car and the truck on the same graph.

Chapter 3 Describing Motion