Unit 2

Unit 2

The Kinematics Equations

(1D Equations of Motion)

Day #1

Introduction to the Equation of Motion

What are the 5 key equations we have learned thus far?

A) x = xf xi finding a change in displacement

B) v = vf vifinding a change in velocity

C) s = d/tfinding speed (distance/time)

D) v = x/tfinding velocity (displacement/time)

E) a = v/tfinding acceleration (change in velocity change in time)

There are 4 new equations for motion.

Well start with a = v/t

Substitute in v = vf vi

The formula is now a = vf-vi

t

Rearrange to get at = vf - vi

Remember

slope of the line =

a = vf-vi

t

Example #1

A car starting from rest accelerates uniformly to a speed of 75 m/s in 12 seconds. What is the cars acceleration?

First, identify the known values, then plug and chug

Vi = 0

Vf = 75 m/s

t = 12 s

a = ?

*always make a list of your knowns

a = vf-vi or at = vf - vi

t

= 75 m/s 0

12s

=+6.25 m/s2

Example #2

A spacecraft traveling at 1025 m/s is uniformly accelerated at a rate of 105 m/s2 for 9 seconds. What is the final velocity of the spacecraft?

First, identify the known values, rearrange the formula, then plug and chug

Vi = 1025 m/s

Vf = ?

t = 9 s

a = 105 m/s2

a = vf-vi or at = vf - vi

t

Vf= Vi + at

= 1025 m/s + 9s(105m/s2)

= 1970 m/s

The second new formula

Recall: area under the curve on a v/t chart gives us displacement = x (or d)

Consider a random, uniform accl object.

Use the area of a trapezoid

Example #3

A plane is moving at a speed of 500m/hr when it lands on a runway. Accelerating uniformly, it comes to a stop after covering 15m. How long did it take to stop?

Vi = 500 m/s

Vf = 0

t = ?

a = ?

x = 15 m

x =t(vf+vi)

15 m = (t)(0+500 m/hr)

30 m = 500m/hr(t)

t = 0.06 hr

= 216 sec

Example #3

A plane is moving at a speed of 500m/hr when it lands on a runway. Accelerating uniformly, it comes to a stop after covering 15m. How long did it take to stop? What was its acceleration?

Vi = 500 m/hr

Vf = 0

t = 0.03 hr or 108 s (previous slide)

a = ?

x = 15 m

a = vf-vi

t

= 0-500m/hr

0.06 hr

=-8333.33 m/hr2

Or -0.0006 m/s2

HW

In book,

Complete pg 53,55, & 58 odds

Day #2

RE-Introduction to the equation of motion

. area under curve =

2.

You have to use the quadratic formula to solve the problem.

The first answer doesnt make sense because you cant have a negative time.

Substitute our newest formula into the derivation below

a = vf-vi (rearrange)

t

List what you have (using COMPATIBLE units & using proper SIGNS)

choose your equation

Solve the equation

Make sure your answer makes sense (both in MAGNITUDE & DIRECTION)

Make a decision as to which direction is POSITIVE & which is NEGATIVE

3.

4.

Review Solve

Solve the following:

You roll a ball up an incline at a speed of 4.5 m/s. After 5 sec, the ball is on the way down travelling at a new speed of 1.5 m/s. Find the balls acceleration. (Sketch)

-1.5-4.5-6-1.2 m/s2

5 5

rookie mistake. 1.5-4.5

Find the time to the peak.

-1.2 = 0-4.5-1.2 = -4.5=3.75 sec

tt

How far up does it go?

0 = (4.5)2 + 2(-1.2)d=8.44 m

Day #3

Free Fall

Free Fall

Free fall problems are acceleration problems where we know the acceleration.

For all free fall problems (at the surface of the earth),

a = 9.8 m/s2 (down) the object speeds up by 9.8m/s2 after every second that it falls

When solving:

Make a chart

Use arrows next to variables to show the direction of the vectors

Fill in known variables

Solve for unknowns

Free Fall Problems

Initial Velocity (Vi) will be 0, so our formulas change slightly. No need to memorize these..

a = vf-vi

t

a = vf

t

0 m/s

9.8 m/s

19.6 m/s

29.4 m/s

39.2 m/s

49 m/s

9.8 m/s2

Free Fall Problems

1. A ball is dropped from the top of a bridge, and it takes 8 seconds to hit the water.

A. How tall was the bridge?

B. How fast was the ball going just before it hits the water?

a. height? So find x. b. a = (vf-vi)/t

x = vi t +1/2(a)(t )2 9.8= vf/8

= (9.8)(8) 2 = 78.4 m/s

= 313.6 m downward

*You could solve part B first and then use the equation x=1/2t(vi+vf)

a9.8m/s2 vi0 vf x t 8s

vi= 0

a = -9.8 m/s2

Dx = -20 m

Because the diver is falling DOWNWARD, we can also use a negative sign

a-9.8m/s2 vi0 vf x20m t

If all objects accelerate at the same rate when falling near the earths surface, why do some objects actually hit the ground faster than others? (assuming they are dropped at the same time from the same height)?

AIR RESISTANCE!!

Some objects are narrower (like pencils) and therefore the air resists their falling less.

Some things are wider (like frisbees) and the air provides more resistance and slows down their fall.

NOTICE HOW THE MASS OF THE OBJECTS DOES NOT MATTER!

5.

y

t

t

v

Notice that the object speeds up

Notice that the acceleration is CONSTANT and NEGATIVE the ENTIRE way

Sometimes the variable y is used instead of x to simply show that the object is moving vertically

Day #4

Throw-ups, Come-downs, Throw-downs

Free Fall Problem Throw up

A ball is thrown up from the ground

In this type of problem, the initial velocity is always up. Also, the primary type of problem will ask what is the maximum height, which means vf = 0 m/s.

2. A ball is thrown up from the ground at an initial speed of 20 m/s.

A. what is the maximum height the ball reaches?

B. How long before it gets to the max altitude?

A. x =?B. t =?

Vf2=vi2 +2a xa=(vf-vi)/ t

0=2o2 +2(-9.8) x-9.8=(0-20) t

x = 20.4m t=2.04s

*You can solve for B first then use the formula

x =1/2t(vi+vf)

a-9.8m/s2 vi20 m/s vf0 x t

UP +, DOWN -

a = -9.8 m/s2

v2 = 0 (at the top)

a)

vi = 20 m/s

a = -9.8 m/s2

vf = 0 m/s

To solve for t, use the quadratic formula

1

2

Problem #2

b)

c)

t= 0.66s and 3.42 sec

(going up) (going down)

Throw up/Come down

Assume:

Up+, Down

A =-9.8m/s2

v2= 0 (at the top)

Well start with scenario #3. V3=-V1

3 scenarios 1. 2. 3.

1

2

3

Scenario 3 object is thrown up and caught at the same height

vf = 0 m/s

a = -9.8 m/s2

Dx = 30 m

vf = 0 m/s

a = -9.8 m/s2

v i= 24.25 m/s

Due to the fact that the ball is thrown and caught at the same height

a)

b)

c)

Scenario 2 and part d of our problem

vi= 24.25 m/s

a = -9.8 m/s2

Dx = -1 m (because down is -)

vf2=vi2 +2ax

vf2=24.252 +2(-9.8)(-1)

24.65 m/s down

Or-24.65m/s

1

2

4

3

vi= 24.25 m/s

a = -9.8 m/s2

Dx = -1 m (because down is -)

To solve for t, use the quadratic formula

Only positive times make sense

What if we were solving for the time it was when the object hits the ground?

Day #5

Chase Problems

Chase Problems

A chase problem is a scenario where two objects are involved, and they have the same position at some later time.

They can

Both start at the same place and same time

Start at different places but at the same time

Both start at the same place but at different times

Start at different places at different times

B or D but be moving in different directions

The Key Equation

x=1/2at2+vitx=xf-xi

Substitute the second formula into the first, creating:

x2=1/2at2+vit+xi

This formula will give you the final position for both objects. Since you dont know that final position, set the two formulas equal to each other.

Chase Problem #1

Timmy is running at a constant speed of 6m/s. He sees Susie running 50m in front of him. She is moving at a constant speed of 4 m/s, in the same direction. How long will it take for Timmy to catch Susie?

Notice that while Timmy and Susie have different starting positions (xiT=0m while xiS=50m), they have the same final position (xf) after Timmy catches Susie. . . X

xiT xiS x2

Tim = Susie

1/2at2+vit+xi = 1/2at2+vit+xi

(0)t2+6t+0 = 1/2(0)t2+4t=50

6t = 4t=50

t = 25sec

Chase Problem #2

A continuation of the last problem.

Timmy passes Susie, running at his constant speed of 6m/s. Susie decides to pick up the pace, very gradually. She begins to accelerate at a constant rate of 0.1m/s2. How long will it take her to catch Timmy?

Notice that Timmy and Susie have the same starting position (xiT=0m xiS=0m) and ending position (xf) after Susie catches Timmy.

.x

xit=xisx2

Tim = Susie

1/2at2+vit+xi = 1/2at2+vit+xi

(0)t2+6t+0 = 1/2(0.1)t2+4t+0

6t=0.05t2+4t

6=0.05t+4

t=40sec

Really, you could have made the initial position anything you wanted they cancel out. I decided to choose 0.

Chase Problem #3

A man drops a penny off the top of a 100m tall building. Exactly 1 second later another man throws a nickel downward from the same place as the first man. What is the minimum speed the nickel must be thrown at in order to catch the penny?

Start with the time it takes the penny to hit the bottom using the formula:

x=1/2at2+vit

-100=1/2(-9.8)t2+0t

t= 4.52 sec

penny = nickel (1 sec later)

1/2at2+vit+xi = 1/2at2+vit+xi

(-9.8)(4.52)2+0(4.52)+0 = 1/2(-9.8)(3.52)2+vi(3.52)+0

Vi= -11.19 m/s (neg. because its going down)

+

-

0

Chase Problem #4 really hard

In a strange, yet exciting, crash-test-dummy crash, two cars start by facing each other 1000m apart on a straight road. The first car accelerates from rest with a constant accel of 4m/s2. The second car accel, at a rate of 8 m/s2 for 5 sec but then settles into a constant speed. Find the elapsed time before these two cars collide.

Notice that the dummies have different starting positions (xiA=0, xiB=1000m), but they have the same final position (x2) when they crash.

Start by finding the final velocity of car B by using the following formula:

a = vf-vi-8=(vf-0)/5 so vf =-40m/s

t

1/2at2+vit+xi = 1/2at2+vit+xi

Test Review

Step #1

Write Down What You Have (Look for Key Words)

Coming to a stopStarting from restCoastingMaximum HeightDropped

vf = 0

vi = 0

vi = vf = constant

vf = 0

a = -9.8 m/s2

Slowing DownBrakingSpeeding upAccelerating from rest

a = - __

a = + __

Step #3

Solve the Equation

Step #4

Make sure your answer makes sense

Helpful Tip #1

Choose your Key Points in every problemand do so wisely.

Free-fallThrow up

vi = 0

vf = 0 (at top)

Vertical Problems

1

2

2

1

vi 0

1

2

Throw-downs

Throw up / Come Down (throw and catch at same height)

vf = 0 (at top)

Vertical Problems

2

1

3

v1 = -v3

Throw up / Come Down (throw and catch at different heights)

Use and solve quadratically for t

Dx13 = +

2

1

3

2

1

3

Dx13 = -

Helpful Tip #2

Assign positive and negative to different directions.

Helpful Tip #3

When solving a quadratic equation, do so with minimal effort.

Solving a quadratic equation

Choice A

Choice B

Factoring

Choice C

2nd Trace Zero

(on graphing calculator)

Unlikely on a physics problem

Can be used at constant speeds (a=0) or when accelerating. Awesome Dude!

Chase Problems

Since the two objects (A and B) end up at the same position by the end of the chase, use

But what if

The objects start at different places?

Its already accounted for here and here

But what if

The objects start at different TIMES?

Youll need to use an extra equation relating the two times. Plug this new equation into the long equation above.

Example: tA = tB + 1

Helpful Tip #6

It is always important to remember that when something is thrown up or down (or simply falls), the acceleration at ALL times is constant.

The acceleration of the ball at EVERY point on this red path (When its rising up, when its stopped, when its falling down) is always -9.8 m/s2.

+

-

An object thrown up has a constant acceleration at ALL times.

.the acceleration due to gravity

Objects rise and fall in the same amount of time (assuming no parachute )

t

x

t

v

Constant slope = constant accel.

TONIGHTS HW

Complete Review Worksheet

area

=

1

2

h

(

b

1

+

b

2

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)

(

2

1

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area

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triangle

+

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rectangle

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1

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=

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77

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