The slides used in this video are color coded. If you are experiencing difficulty with one aspect of your understanding than another you might find this coding… useful!Slides with red backgrounds involve word problems.

Slides with tan backgrounds involve matching concepts.

Slides with olive backgrounds involve reading data tables.

Slides with green backgrounds involve graphing.

Dr. Fiala is traveling on hisHarley at a constant 13.67

m/s. What is the distancetraveled by Doc in 7.32seconds?

SOLUTION:Find the distance traveled.

Vi&f = 13.67 m/s Xf

ti = 0 stf = 7.32 sXi = 0 m Xf= ma = 0 m/s2

SOLUTION:Find the distance traveled.

Vi&f = 13.67 m/s Xf

ti = 0 stf = 7.32 sXi = 0 m Xf= 100.07 ma = 0 m/s2

Dr. Fiala notices he is nowtraveling at a constant 49.21 km/h. What is the distancein meters traveled by Doc in 7.32seconds?

SOLUTION:Dimensional analysis.

49.21 km/h = 13.67 m/sSo the Xf remains

100.07 m

X X X

M K H D 0 d c m

1 x x x

Dr. Fiala jumps in his un-started car. He accelerates

at arate of 4 m/s2 for 8 seconds.How far did Doc travel?

vi = 0 m/s vf =ti = 0 s Xf =tf = 8 s a = 4 m/s2

Xi = 0 m Xf = 128 m

Doc’s final position.

If Dr. Fiala starts from a fullstop and accelerates at 2.25 m/s2, how incredibly fastwill he be traveling when he has traveled 1530 meters?

SOLUTION:Calculate final velocity without knowing time.

Vi = 0 m/s Vf

Xi = 0 m tf

Xf = 1530 ma = 2.25 m/s2 ti = 0 s

Vf= 82.98 m/s

If Dr. Fiala starts from a fullstop and accelerates at 2.25 m/s2, how long will it

takehim to drive 1530 meters?

SOLUTION:Solve for time.

Vi = 0 m/s tf

Xi = 0 m Vf

Xf = 1530 ma = 2.25 m/s2 ti = 0 sVf = 82.98 m/s

tf= 36.88 s

2.25 + 3.25 =

2.25 + 3.25 =

2.25 + 3.25 =

SOLUTION:Adding vectors.

+2.25 + +3.25 = +5.5

+2.25 + -3.25 = -

1.00

+2.25 + +3.25 = +3.95

Determine the mass of a 153.08 N

object.

153

N

?

kg

SOLUTION:Calculate mass.

W = 153.08 N mg = -9.8 m/s2

m = 15.62 kg

Determine the force of friction on a

15.62 kg object traveling at aconstant horizontal velocity of3.62 m/s while experiencing anapplied force of 6 N.

SOLUTION:Calculate force of friction.

m = 51.62 kg Ff

a = 0 m/s2 g = -9.8 m/s2

Fa = 6 N Ff = -6 N

Force Diagra

msMotion

Map

Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?

Determine the force needed toaccelerate Dr. Fiala’s car and itsoccupants at a rate of 3.23 m/s2 ifthe total mass of car and occupantsis 1315 kg and there is no frictionforce.

SOLUTION:Find applied force.

m = 1315 kg Fa = 3.23 m/s2

F = 4247.45 N (kg)(m/s2)

This time, when we apply that4247.45 N force to Dr. Fiala’s car andits occupants, the resultingacceleration is actually lower. Itregisters at a rate of only 3.00 m/s2. What is the magnitude for the forceof friction causing the acceleration to

bedecreased?

SOLUTION:Find applied force.

m = 1315 kg Ff

F = 4247.45 Na = 3.00 m/s2

Ff = - 302.45 N

When an object is freefalling it isweightless. Prove mathematically

thata .448 kg apple is weightless during

itsfreefall from a tree. Draw a forcediagram of the apple during its fallfrom the tree.

SOLUTION:Find force of support.

m = .448 kg Fs

g = -9.8 m/s2

Fs = 0 N

Fg = -4.39 N

Assuming a perfectly frictionless surface, ideal forlaunching students in a game of faculty bowling,Dr. Fiala uses a brand new gizmo that automaticallyapplies a force that results in an acceleration of 1.1 m/s2. Experimentation resulted in a student with

amass of 44.10 kg, accelerating at 1.1 m/s2. Find theforce generated by the gizmo for that student.

SOLUTION:Find force in the horizontal.

m = 44.10 kg Fa = 1.1 m/s2

F = 48.51 N

All of the students from theprevious problem (combined mass) step into an elevator at the same time. Draw a force diagram of this situationincluding the magnitude of Fg and Fs.

SOLUTION:Find force of gravity and force of support.

m1 = 42 kg Fg

m2 = 43.25 kg Fs

m3 = 44 kgm4 = 44.23 kgm5 = 44.77 kgm6 = 45.01 kgm2 = 45.45 kg Fg= -3025.36 Ng = -9.8 m/s2 Fs= 3025.36 N

This same elevator accelerates

at a rate of .75 m/s2 towards the

second floor. Draw a force

diagram of this situation

including the magnitude of Fg and

Fs.

SOLUTION:Find force of support.

m = 308.71 kg Fs

Fg = -3025.36 N g = -9.8 m/s2

a = .75 m/s2

Fs= 3256.89 N

Fg = 3025.36 N

Fs = 3256.89 N

This same elevator acceleratesat a rate of .50 m/s2 as it begins

itsstop for the second floor. Draw aForce diagram of this situationincluding the magnitude of Fg

and Fs.

SOLUTION:Find force of support.

m = 308.71 kg Fs

Fg = -3025.36 N g = -9.8 m/s2

a = .-50 m/s2

Fs= 2871.01 N

Fg = 3025.36 N

Fs = 2871.01 N

According to Newton’s 3rd law, anaction force causes an equal on oppositereaction force. It is no wonder a truckwindshield squashes a bug and not viceversa. A 2000 kg truck and a .0002 kgbug hit with a 50 N force. Take a closerlook at why the truck wins the collisionby calculating the accelerationexerienced by the bug and by the truck.

SOLUTION:Why the bug doesn’t survive.

mt = 2000 kg at mb = .0002 kg ab

g = -9.8 m/s2

F = -50 N at = -.025 m/s2

ab = -250,000 m/s2

These cables will snap if themass of the trafffic light exceeds10.1 kg. Does the traffic lightexceed 10.1 kg?

375.40

N

SOLUTION:The cable does not break.

T1 = 375.4 N mg = -9.8 m/s2 T1y

Θ = 7.5° m= 10 kg

Dr. Fiala attempts to walkdue east at 5 m/s at thesame time as a 30 m/s cold,winter wind is blowing duesouth. What is themagnitude of Dr. Fiala’s

velocity.

SOLUTION:Resultant velocity magnitude.

Vi = 30 m/s Vf a2

+ b2

= c2

Vi = 5 m/s

Vf= 30.41 m/s

Vy = 30 m/s

Vx = 5 m/s

If Dr. Fiala continues hisvelocity and the windcontinues to blow steadily,at what angle, as measured

frompositive “X”, is Dr. Fiala’svelocity.

Vx = 5 m/s

Vy = 30 m/s

V = 30.41 m

/s

SOLUTION:

Vy = 30 m/s

Vx = 5 m/s

tan Θ = x yΘ = 9.46°

tan Φ = y x

Φ = 80.54°

Θ (from +x) = 279.46°

Resultant velocity angle measured from positive x.

Because of this wind, a 15 kgpackage is blown from Dr. Fiala’sarms and onto the ground. The 15 kgpackage reaches a velocity of 30.41 m/s in a time of 4 seconds. Find theforce acting on the box horizontally ifthere is no friction.

SOLUTION:Find applied force.

Yf = -15 m aYi = 0 m Fm = 15 kgg = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/sti = 0 s a = 7.85 m/s2

ti = 4 s F = 117.79 N

If the package is blow horizontally at30.41 m/s off a ledge onto a parkinglot that is 15 meters below how muchtime will it spend in the air beforestriking the ground? What does themotion map look like?

SOLUTION:Find time package spends in the air.

Yf = -15 m Vf

Yi = 0 m tf

m = 15 kgg = -9.8 m/s2 tf = 1.75 sVi = 0 m/s ti = 0 m/s

Dr. Fiala throws a baseball in theair with an initial velocity of 27 m/s atan angle of 27° to the horizon. Create avelocity vector diagram and show, byparallelogram method, the “X” and “Y” components of the baseball’s velocity.

SOLUTION:Resolve velocity vector into “x” and “y” components just like force or any other vector.

V = 27 m/s Viy

Θ= 27° Vix

g = -9.8 m/s2 Viy = 12.26 m/s Vix = 24.06 m/s

Vx = V

Vy = V

27 m/s

27°

How much time will it take for thebaseball to reach the same heightfrom which it was thrown?

SOLUTION:Find time in the air.

g = -9.8 m/s2 tf

Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/sti = 0 sYi = 0 mYf = 0 m tf = 2.5 s

How far will the baseball travel in2.5 seconds?

SOLUTION:Find range.

g = -9.8 m/s2 Xf

Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/sti = 0 sYi = 0 m Yf = 0 mXi = 0 m tf = 2.5 s Xf = 60.15 m

What is the maximum height the

baseball attained during its flight?

SOLUTION:Find Δy.

g = -9.8 m/s2 ΔyΘ= 27° Viy = 12.26 m/s Vix = 24.06 m/sti = 0 sYi = 0 m Yf = 0 mXi = 0 m Δy = 7.67 mtf = 2.5 s Xf = 60.15 m

If it was a .448 kg apple that wasthrown into the air at 30 m/s whatwas the apple’s intial momentum?

SOLUTION:Find momentum of apple.

m = .448 kg pVi = 30 m/sg = -9.8 m/s2

p = 13.44 kgm/s

What constant force is needed toget a change in the apple’smomentum from 13.44 kgm/s to 0In 3.06 seconds?

SOLUTION:Find force necessary to change momentum.

m = .448 kg FVi = 30 m/sg = -9.8 m/s2

ti = 0 stf = 3.06 sΔp = -13.44 kgm/s

F = 4.39 N

After falling to the ground the .448 kg apple rolled at a constant10.4 m/s where collided with astationary .577 kg apple. If the

twoapples stuck together, at whatvelocity would they roll?

SOLUTION:Find the velocity of two apples stuck together.

m1 = .448 kg Vf

m2 = .577 kg pg = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s

p = 4.66 kgm/s2

Vf = 4.55 m/s

Determine the force applied ifthe rolling apples strike a walland a come to a stop in .311seconds.

SOLUTION:Find force needed to stop apples.

m1 = .448 kg Fm2 = .577 kgti = 0 stf = .311 sg = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s F = 14.98 N