chapter 5 motion

Chapter 5 MotionChapter 5 Motion

Average VelocityTimeTotal

DistanceTotal

if

if

tt

dd

t

d

V

if

if

tt

vv

t

v

a

if t1 = 0,

t

vv if a

d = v t


t

vv if a Solve for vf

atvv if


tvv

d fi

2

but 2

fi vvv

tvd


tvv

d fi

2but atvv if

tv

d i

2

atvi tatv

d i

2

2

2

2

1attvd i


tvv

d fi

2atvv if

Solve both equations for t

fi vv

dt

2a

vvt if


fi vv

dt

2

a

vvt if

Set equal to each other

fi vv

d2

a

vv if


fi vv

d2

a

vv if Multiply across

222 if vvad

advv if 222


advv if 222

2

2

1attvd i

d = v t

t

vv if a

tvv

d fi

2

atvv if


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



vi = 18 m/st = 3 svf = 0s

sm

t

vva if

3

/180

2/6 sma



2

2

1attvd i

22 )3)(/6(2

1)3)(/18( ssmssmd =27 m


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

ma

vvd if 126

2

22


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/sa

vv if t

ssm

smsm58.4

/12

/0/552

t


Scott 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


a

vvd if

2

22

)/3(2

)/20(02

2

sm

smd

md 7.66

The cowsurvives!!!!

Acceleration Due to Gravity

ag = -9.8 m/s2

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


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?


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

vf = 0a

vv if t 2/8.9

/490

sm

sm

t

sec5t Total time in air is 10 sec


How high does it go?

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

t = 5 sec

2

2

1attvd i

22 )5)(/8.9(2

1)5)(/49( ssmssmd

md 5.122


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

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

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.

Stoplight Problem

a. How far will the car travel before it overtakes the truck?

• Truck d = vt Car d = vit + ½ at2

• d = 21 t d = ½ (6) t2

• 21t = 3t2

• 3t2 – 21t = 0• 3t(t – 7) = 0 t = 0 s, 7 s

Stoplight Problem

b. How fast will the car be traveling when it overtakes the truck?

•d = 21 t = 21(7) = 147m •d = ½ at2 = (1/2)(6)(7)2 = 147m

c. Construct a distance vs time graph for the car and the truck on the same graph.

• Graph y1 = 21t, y2 = 3t2 with the –0 symbol. Under mode select simultaneous with a window of [0,10],[0,150]

Stoplight Problem

chapter 5 motion

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