MechanicsMotion in One DimensionVectors and Two Dimensional Motion

Chapter 1 (p.2-22)Chapter 2 (p.23-50)Chapter 3 (p.51-78)

Basic ConceptsProblem solving strategy:1.Read problem2.Draw diagram3.Identify data4.Choose equation(s)5.Solve6.Evauate and check answer

SI Units:Length=meterMass=KilogramTime=Second

*** Remember significant digits ***

Key Equations

Displacement:an objects change in position given by final minus initial coordinates.

Average VelocityV = x/t

Instantaneous Velocity V = lim x/t t→0

Acceleration

• Average Acceleration:a = v/t• Instantaneous Acceleration: of an object at a

certain time equals the slope of a velocity-time graph at that instant.

• One dimension motion with constant acceleration:

- V = vo+at

- V = (vo+v)/2

- X = vot+.5at2

- V2 = vo2+2ax

Projectile Motion

Problem Solving Strategy

1. Select a coordinate system2. Resolve the initial velocity vector into x

and y components 3. Treat the horizontal motion and the

vertical motion independently4. Analyze the horizontal motion of the

projectile with constant velocity5. Analyze vertical motion of the projectile

with constant acceleration

Equations for Projectile Motion

x = vxot

vy = vyo-gt

y = vyot-.5gt2

vy2 = vyo

2-2gy

g = 9.8 m/s2 (on earth…but you can do projectile motion on other planets if you know the local value of g. Also, some electric field problems can be solved in the same manner as projectile problems!)

Relative VelocityDifferent displacements and velocities may be

observed for objects in motion depending on the frame of reference.

For example: A mouse is running at 1.2 m/s backwards on a canoe that is moving forward at 3.4 m/s on a calm pond.

Vmc= -1.2 m/s (velocity of mouse relative to canoe)

Vcp= +3.4 m/s (velocity of canoe relative to pond)

Vmp=vmc+vcp=+2.2 m/s (velocity of mouse relative to the pond)

A physical quantity that requires both direction and magnitude

VectorVector

Has only magnitude and no direction

Scalar quantityScalar quantity

Determine which of the following quantities are vectors, and which are scalars.MomentumSpeedkinetic energyWorkAccelerationForcePowerElectric fieldelectric potentialelectric chargeTorquemagnetic fieldTemperatureTimeDisplacementdensity

VectorScalarScalarScalarVectorVectorScalarVectorScalarScalarVectorVectorScalarScalarVectorScalar

_______ is described as the limit of the average velocity as the time interval becomes infinitesimally short.

Instantaneous Velocity

The average velocity on a position-time graph is the _____ of the straight line connecting the final and initial points.

slope

The ______ is zero in a projectile problem when the object is traveling through the air and is acted on only by gravity.

A) Vertical Velocity C) Horizontal Velocity

B) Horizontal Acceleration D) Speed

B) Horizontal Acceleration B) Horizontal Acceleration

The ______ is zero in a projectile motion problem when the object reaches the highest point of its trajectory.

A) Speed C) Acceleration

B) Horizontal Velocity D) Vertical velocity

D) Vertical Velocity D) Vertical Velocity

The 3 equations that describe the motion of an object moving with constant acceleration along the x-axis are:

V = vo+at

X = vot+.5at2

V2 = vo2+2ax

Also… x= ½ (v+vo)t

A race car starting from rest accelerates at a rate 5 m/s2. What is the velocity of the car after it has traveled 100 ft?

V2 = vo2 + 2ax

V2 = (0)2+2(5m/s2)(30.5m) = √(305m2/s2) = ± 17.5m/s

A golf ball is released from rest at the top of a really tall building. Neglecting air resistance, calculate the position and velocity of the ball after 2 seconds.

V = at = (-9.8m/s2)(2s) = V = at = (-9.8m/s2)(2s) = -19.6m/s-19.6m/sY = .5at2 = .5(-9.8m/s2)Y = .5at2 = .5(-9.8m/s2)(2s)2 = (2s)2 = -19.6m-19.6m

A passenger at the rear of a train, traveling at 15m/s relative to the earth, throws a ball with a speed of 15m/s in the opposite direction of the trains motion. What is the balls velocity relative to the earth?

vvtete = 15m/s v = 15m/s vbtbt = -15m/s = -15m/svvbebe = v = vbtbt + v + vtete = 0 = 0

Adding vectors:1) Select a coordinate system 2) Draw a sketch and label vectors3) Find x and y components of all 4) Find the resultant components in both directions5) Use ______ to find the magnitude of the resultant vector

Pythagorean theoremPythagorean theorem

The _____ acceleration of an object at a certain time equals the slope of a velocity-time graph at that instant.

instantaneousinstantaneous

The average velocity of an object moving along a line can be either positive or negative depending on the sign of the displacement.

True or False

TrueTrue

What is the SI unit for volume?

mm3 3

The slope of a line tangent to the position-time curve at a point is equal to the ________ of the object at that time.

Instantaneous velocity Instantaneous velocity

The displacement of an object during a time interval can be found by determining the _________ on a velocity-time graph.

Area under the curveArea under the curve

What does the area under the curve of an acceleration time graph tell you about the motion of an object?

How much its velocity How much its velocity changed during that time changed during that time interval.interval.

The velocity of a car decreases from 30m/s to 10m/s in a time interval of 2 seconds. What is the acceleration?

A = A = 10 – 3010 – 30 2.02.0

A = -10m/sA = -10m/s22

Which equation represents velocity as a function of displacement?

VV22 = v = voo22+2ax+2ax

Which equation represents displacement as a function of time?

X = vX = voot+.5att+.5at22

A plane traveling at 4m/s at a height of 100m above the ground drops a package. Where does that package fall relative to the point at which it was dropped?

Horizontal: x = vHorizontal: x = vxoxot =(40m/s)tt =(40m/s)t

Y = -.5gtY = -.5gt22

-100m=-.5(9.8)t-100m=-.5(9.8)t22

T=4.51sT=4.51s

X=(40)(4.51)=X=(40)(4.51)=180m180m

A stone is thrown upward from the top of a building at an angle of 30º to the horizontal and with an initial speed of 20m/s. The height of the building is

45m. How long is the stone in flight? Where does it land? At what angle does it strike the ground?

vxo=vocosθo=17.3m/s vyo=vosinθo= 10m/sY=vyot-.5gt2

-45m=(10)t-.5(9.8)t2

T=4.22sX=vxot= 73m Vy=vyo-gt= -31.36 m/sVx=vox=17.3 m/s Tan=vy/vx =61o

below horizontal

A golf ball is released from rest at the top of a tall building. Neglecting air resistance calculate the position and velocity of the ball after 3 seconds.

V=at=(-9.8)tV=at=(-9.8)tV=(-9.8)(3)= V=(-9.8)(3)= -29.4m/s-29.4m/sY=.5atY=.5at22=.5(-9.8)t=.5(-9.8)t22

Y=.5(-9.8)(3)Y=.5(-9.8)(3)22= = -44.1m-44.1m

A stone is thrown from the top of a building with an initial speed of 20m/s straight upward. The building is 50m high and the stone just misses the edge of the roof on the way down. Determine a) the time for the stone to reach its max height, b) the max height, c)the time needed for the stone to return to the level of the thrower.

a)20m/s + (-9.8m/s2)t=0 t=2.04s

b) ymax=(20m/s)(2.04s) + .5(-9.8m/s2)(2.04)2

ymax= 20.4m

c) y=vot + .5at2

0=20t – 4.90t20=t(20-4.90t)t=0 t=4.80s

Explain what each of these equations are used for in a projectile motion problem involving a plane dropping a package to the ground.

a) y= -.5 gt2 b) Vy= vyo- gt c) x=vxot

a)This equation can be used to determine the time of flight (initial vertical velocity is zero for a dropped object)

b) This equation is used to find the vertical component of the velocity just before the package hits the ground.

c)Since the package is dropped from the plane it has a horizontal velocity and moves horizontally at a constant rate (zero acceleration) the entire time it is in the air.

What equation is used to find the maximum height in a projectile motion problem?

And in what column, horizontal or vertical, should this equation solved?

The max height is found using y= (vosinθo)t - .5gt2 and it should be in the vertical column.

A stone is thrown upwards from the top of a building at an angle of 30º to the horizontal and with an initial speed of 20m/s. The height of the building is 45m.

a) How long is the stone in fight?b) What is the speed of the stone just before it strikes the

ground?

Vxo= vocosθ= +17.3m/s Vyo= vosinθ = +10m/sy= vyot- .5gt2

-45m= (10m/s)t- .5(9.8)t2

T= 4.22svy= vyo- gt= -31.4m/s

v=√(vx2+ vy

2)= 35.9m/s

see slide #30 to find how to calculate horizontal distance, or angle of final velocity vector.

Explain all variables in this equation? Vy

2=vyo2-2g

Vy2 = final velocity in the y

directionVyo

2 = initial velocity in the y directiong = downward acceleration due to gravity (9.8m/s2 on earth )

Define instantaneous velocity and give its equation.

Instantaneous velocity is Instantaneous velocity is defined as the limit of the defined as the limit of the average velocity as average velocity as t goes t goes to zero. to zero. v= lim v= lim r/r/tt

tt→→00

Define average velocity and give its equation.

Average velocity of an object moving Average velocity of an object moving along the x-axis during some time along the x-axis during some time interval is equal to the displacement interval is equal to the displacement of the object divided by the time of the object divided by the time interval over which the displacement interval over which the displacement occurred.occurred.V= V= x/x/tt == (x(xff-x-xii)/(t)/(tff-t-tii))

A car traveling at a constant speed of 30m/s (67mi/h) passes a trooper. One second later, the trooper sets off to chase the car with a constant acceleration of 30m/s2. How long does it take the trooper to overtake the car?

Car Trooper a=zero a=30m/s2 (!!)xo=30 meters xo=0 metersv=30m/s (constant ) vo=0m/sxc=xo+vt xt= .5at2

Set their positions equal and solve for time:

xo + vt = .5at2 t=[v±(v2+2axo)½]/a t=2.73s