Physics 101

Prof. Ekey

Chapter 2

Kinematics in one dimension

Uniform motion, s vs t, v vs t, a vs t, kinematic equations… fun.

“In this chapter, you will learn how to solve problems about motion along a straight line”

w.r.t.= with respect to

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slope = m =ΔyΔx

=(y2 − y1)(x2 − x1)

=riserun

Math review Graphing Lines – (x,y) coordinate system

y-intercept = b = value of y when x =0.

For any curve find slope by taking derivative w.r.t. x, and plug in point of interest… or… just use your pencil (more on this later)

Match the line to the equation.

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y = mx + b

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y = x +1

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y = −x +1

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y = 12 x −1

x

y

5

5

-5

-5

€

vavg ≡ΔxΔt

or ΔyΔt

= slope of position vs. time graph

“Straight-line motion in which equal displacements occur during any successive equal-time intervals.”

Average velocity is the slope of a position vs. time graph. “An object’s motion is uniform if and only if its position vs. time graph is a straight line” & “vx and vy are constant and unchanging”

Time [s]

Posit

ion

[m]

0

4

-4

2 4 6 8

Is this Uniform Motion? Which has a larger velocity? Larger Speed? When do they meet? What happens at 4.0 s? Do either change direction?

Uniform Motion – constant velocity, no acceleration

Questions to ask: Where and when did it start and stop? What direction is the movement? Is the slope (vel) + or -? How does this slope compare?

Time [s] Posit

ion

[m]

Time [s] Vel

ocity

[m/s]

0

0

4

-4

2 4 6 8

4

-4

Describe the motion Calculate the slopes (velocities) Graph your results on v vs. t

2 4 6 8

A

B

C

Velocity vs. time for uniform motion

vA =ΔxΔt

=4.0m−1.0m3.0s− 0.0s

=1.0ms

7.5 km

finish Bird 30 km/hr

15 km/hr

(a)

(b)

(c)

(d)

Questions

A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to the finish at 30 km/hr. When the bird reaches the finish line, it turns around and flies back to the runner, and then turns around again, repeating the back-and-forth trips until the runner reaches the finish line. How many kilometers does the bird travel? (a) 10 km (b) 15 km (c) 20 km (d) 30 km Which position-versus-time graph represents the motion shown?

Why is this an ideal situation?

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v = limΔt→0

ΔsΔt

=dsdt

Book uses “s” to denote straight-line motion in the x, y or any other straight-line direction.

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s = t 2

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s = 2

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s = ctn

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s = 2t

Instantaneous velocity Velocity (Speed and direction) at a particular instant in time - speedometer plus a compass

“the instantaneous velocity at time, t, is the average velocity during a time interval, ∆t, centered on t as ∆t approaches zero.”

“the instantaneous velocity at time t is the slope of the line that is tangent to the position vs. time graph.” Derivatives (review and/or introduction) Take the derivative w.r.t. time for the following functions

Non-uniform motion (standard form of motion) Average velocity

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v = ΔxΔt

=8.0m − 0.0m8.0s − 0.0s

=1.0 ms

Instantaneous Velocity Speedometer + compass Tangent line to the curve at the time of interest

Describe the motion. Direction traveled? Compare inst. vel. with avg. vel.

Use your pen as a slope-measuring device.

Time [s]

Posit

ion

[m]

0

8

2 4 6 8 C

A D B

E

A

D

E

B

C

Cars, Trucks and Buses A car’s position as it accelerates is modeled by the equation s = 2t2 [m],where t is in seconds as shown in the graph Describe the motion (s, v & a) What is the car’s location at t= 2.0 seconds? Using the graph and/or a derivative, determine the car’s velocity at t=2.0 seconds? What’s the car’s initial velocity? Is the car accelerating?

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s f = si + vsdtti

t f

∫

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s f = si +

Time [s]

Vel

ocity

[m/s]

Time [s]

0

2 4 6 8

4

-4

A

B

C

2 4 6 8

Posit

ion

[m]

0

4

-4

∆ sB = Area of  box =  ∆ vB( ) ∆ tB( ) = 1.0 m / s ( ) 3.0 s( ) = 3.0m

Position from v vs. t Requires integration of vs(t)

or calc. area under v vs. t curve.

Calculate ∆s for A & C. What’s your total displacement? Graph s vs t.

What’s your starting position?

Area under vs from ti to tf

Integrate (area)

Differentiate (slope)

Slope of velocity vs time

Straight-line motion with constant acceleration “An object has uniformly accelerated motion if and only if its acceleration as is constant and unchanging. The object’s velocity versus-time graph is a straight line and as is the slope of that line. ” For the graphs, what is the initial velocity?

If after 2.0 seconds the object has a velocity of 8.0 m/s, what is its acceleration?

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a s =Δ v sΔt

=d v sdt

=

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t =−b± b2 − 4ac

2a€

at 2 +bt + c = 0

Kinematic Equations (constant acceleration)

When doing these problems ask yourself.

What is asked? What is given? What can you assume? Do I need to convert any units? starting position, initial time Which equation(s)? Is my answer reasonable?

A pictorial representation can/will/may help. Use x, y in your problem instead of the generic “s”. Replace i and f with numbers when appropriate

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s f = si + visΔt + 12 as (Δt)

2

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vfs = vis + asΔt

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vfs2 = vis

2 + 2as (Δs)

Derived from definitions of ∆s, v and a See the book for the derivations

You may want to do (b) before you (a) or find “v”. Unless you like solving the quadratic equation

Watch for v=0 during motion Uniform motion has as=0

And they’re off. You and a friend are racing your bikes on a level bit of road. You both start from rest at the same initial location. After 10 seconds you cross the finish line and you’re moving with a velocity of 5.0 m/s. What’s your acceleration? How far away is the finish line from the starting point? Your friend pedaled with a constant acceleration of 1.0 m/s2. What’s your friend’s velocity when they cross the finish line? Whoever won, slams on their brakes applying an acceleration of -20 m/s2, so they can jump of their bike and do a happy dance. How long did it take to stop?

See the city. See the zoo. Traffic light won’t let me through. You are driving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car breaks with constant acceleration.

Draw a pictorial representation. How far are you from the intersection when you begin to apply the brakes? What acceleration will bring you to rest right at the intersection? How long does it take you to stop after the light turns red?

SOP sign A car starts from rest at a stop sign. It accelerates at 4.0 m/s2 for 6.0 s, coasts for 2.0 s and then slows down at a rate of 3.0 m/s2 for the next stop sign.

Distance traveled during the 4.0 m/s2 acceleration:

Determine the final velocity after the 4.0 m/s2 acceleration (useful to ans next ques) Determine the distance traveled during the coast Determine the distance traveled during the -3.0 m/s2 acceleration

How far apart are the stop signs?

#2.19

Velocity vs. time for constant acceleration

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Δv = as ⋅ Δt ⇒ vfs = vis + asΔt

Time

Vel

ocity

0

vi

vf

vi

vf = vi+ at

at

t

Time

Vel

ocity

0

vf vi

vf = vi - at

-at

t

vi

Motion in positive direction Positive increasing velocity constant positive acceleration

Motion in positive direction Positive decreasing velocity constant negative acceleration

In these graphs vf, vi and a are magnitudes, the direction is indicated by the ±

Acceleration is the slope of velocity vs time

Velocity vs. time for constant acceleration

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Δv = as ⋅ Δt ⇒ vfs = vis + asΔt

Time

Vel

ocity

0

-vi

-vf

-vf = -vi - at

t

-at

-vi

Time

Vel

ocity

0

-vf

t

vi

-vf = +vi - at

-at vi

Motion in negative direction Negative increasing velocity constant negative acceleration

Motion switches from positive to negative direction Positive decreasing velocity switches to negative increasing velocity constant negative acceleration

In these graphs vf, vi and a are magnitudes, the direction is indicated by the ±

Acceleration is the slope of velocity vs time

Acceleration: velocity vs. time Constant uniform acceleration.

Time [s]

Vel

ocity

[m/s]

Time [s] Acc

el. [

m/s2 ]

0

0

4

-4

2 4 6 8

4

-4

Describe the velocity Calculate accel for B and C. Graph your results on a vs. t

A

B

C

2 4 6

D

8

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aA =ΔvΔt

=4.0m / s − 0.0m / s2.0s − 0.0s

= 2.0m / s2

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aD =ΔvΔt

=0.0m / s − (−4.0)m / s

8.0s − 6.0s= 2.0m / s2

x, v and a vs. t

Time [s]

Vel

ocity

[m/s]

Time [s]

Acc

el. [

m/s2 ]

2 4 6 8

2 4 6

Time [s]

Posit

ion

[m]

2 4 6 8

A

B

C

D

8

Describe the motion Sketch v vs. t. Sketch a vs. t. (straight lines only) Is this motion ideal?

Question.

The graph shows position as a function of time for two trains running on parallel tracks. Which is true: (a) At time tB both trains have the same velocity (b) Both trains speed up all the time (c) Both trains have the same velocity at sometime before tB (d) Somewhere on the graph, both trains have the same acceleration

Time Po

sitio

n

A

B

tB

Gravity (the weight of it all) Free-fall acceleration/Acceleration due to gravity, g = 9.80 m/s2

Why? Gravitational attraction between the mass of the earth and mass of the stuff on the surface. Is “g” mass dependent? Is this a base quantity in SI units? Why? Direction of gravity? What is “a” for gravitational problems? Should we worry about air resistance? Do we need to rewrite our kinematic equations?

Magnitude

If something is moving downward is vy + or -?

Difference Between Indifference and Ignorance

Joe forgot his guitar. He yells to Jim on the 3rd floor to bring down his guitar to the street. Jim being a good and helpful friend drops it out the window. Joe is on the ground directly below the window, and catches the guitar 1.0 m from the ground.

Jim

Joe

15 m

What is the velocity of the guitar when he catches it? How long did it take to fall?

What else could I ask?

Questions

You throw a ball straight up in the air. What can you say about the ball’s acceleration at the top of its motion? (a) zero. (b) less than the acceleration due to gravity. (c) equal to the acceleration due to gravity. (d) more than the acceleration due to gravity. A ball is thrown downward (not dropped) from the top of a tower. After being released its downward acceleration will be. (a) zero. (b) less than the acceleration due to gravity. (c) equal to the acceleration due to gravity. (d) more than the acceleration due to gravity.

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a|| = ±gsinθ

+y

+x

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a free− fall = a || + a ⊥

+y

+x

Convenient to tilt the x/y axis, so x-axis points up (or down) the ramp/incline. What happens if q =0º? 90º?

Motion on an inclined plane (ignore friction) Acceleration is a piece of the free-fall acceleration (g) The correct direction depends on the direction of tilt of the ramp.

Not another car problem. A car traveling at 30 m/s runs out of gas while traveling up a 20º slope. Draw a picture of the car traveling up the ramp. How far up the hill will the car coast before starting to roll back down? How long did this take? After 10 seconds, what is the cars velocity? What happened to the car?