Turn in your homework in the front. Begin: Journal 9/03

1. Write the equation for distance using time and velocity.

2. Write the equation for velocity using time and acceleration.

3. Calculate the acceleration of a runner that can go from rest to 3 m/s in 2s.

4. How long does it take the runner to travel 100 m if he is running at an average speed of 2.7 m/s?

5. Sketch the graphs below. Which position time curve is impossible? Why?

Turn in your homework in the front. Begin: Journal 9/03

1. Write the equation for distance using time and velocity.

2. Write the equation for velocity using time and acceleration.

3. Calculate the acceleration of a runner that can go from rest to 3 m/s in 2s. a = v/t = (3m/s)/2s = 1.5 m/s2

4. How long does it take the runner to travel 100 m if he is running at an average speed of 2.7 m/s?

t=d/t = 100m /2.7m/s = 37.03 s

5. Sketch the graphs below. Which position time curve is impossible? Why?

Physics

More Linear Motion!!!!

Reminder: Problem Solving Strategies

• Memorize the variables that are given in each equation!!!!– example: v = velocity

• Know the units for a given variable– example: velocity is measured in units of

distance over time (meters per second or m/s)– Hint: anytime you see a number in a word

problem, write the variable next to it!

• When reading a word problem identify what you are given– These are your “knowns”

• Identify what you are solving for– This is your “unknown”

Uniformly Accelerated MotionThrough derivations (a fancy way of saying use

algebra to rearrange your problem ) we can rewrite

a = v/t in two new ways:

v = v0 + at

x = x0 + v0t + 1/2 at2

V0 is the initial velocity

X is the final distance

x0 is the starting point (initial distance)

Remember: v = d or x t t

Example Problem

• A skateboarder traveling at 1 m/s accelerates at 1.5 m/s2 for 3 seconds. What is the skateboarder’s velocity after 4 seconds?

v = v0 + at

Example Problem

• A skateboarder traveling at 1 m/s accelerates at 1.5 m/s2 for 3 seconds. What is the skateboarder’s velocity after 4 seconds?

v = v0 + at

v = 1 m/s + (1.5 m/s2)(3 s)

v = 5.5 m/s

Journal Practice

A car traveling at 5 m/s accelerates at 3 m/s2 for 8 seconds. What is the car’s velocity after 8 seconds?

Journal Practice

A car traveling at 5 m/s accelerates at 3 m/s2 for 8 seconds. What is the car’s velocity after 8 seconds?

v = v0 + at

Journal Practice

A car traveling at 5 m/s accelerates at 3 m/s2 for 8 seconds. What is the car’s velocity after 8 seconds?

v = v0 + at

v = (5 m/s) + (3 m/s2)(8) = 29 m/s

Example• A car traveling at 8 m/s accelerates at 5 m/s2

for 10 seconds. • How far has the car traveled 10 seconds after

it began accelerating?

x = x0 + v0t + 1/2 at2

Example• A car traveling at 8 m/s accelerates at 5 m/s2 for 10

seconds. • How far has the car traveled 10 seconds after it

began accelerating?

x = x0 + v0 t + 1/2 at2

x = x0 + (8 m/s) (10 s) + 1/2 (5 m/s2)(10 s)2

Set initial distance or starting point as 0.

x = 0 m + (8 m/s) (10 s) + 1/2 (5 m/s2)(10 s)2

X= 330 m

Journal PracticeA car traveling at 5 m/s accelerates at 3 m/s2 for

8 seconds. • How far has the car traveled 2 seconds after

it began accelerating?

x = x0 + v0t + 1/2 at2

Journal PracticeA car traveling at 5 m/s accelerates at 3 m/s2 for 8

seconds. • How far has the car traveled 2 seconds after it

began accelerating?

x = x0 + v0t + 1/2 at2

t = 2 seconds

x0 = initial distance (we can set our initial distance as zero)

Plug and Chug!!!

Journal Practice# 7 A car traveling at 5 m/s accelerates at 3

m/s2 for 8 seconds. • How far has the car traveled 2 seconds after

it began accelerating?

x = x0 + v0t + 1/2 at2

X = 0 + (5 m/s)(2 s) + (1/2)(3 m/s2)(2 s)2

X = 16 m

Journal

A car accelerates from rest at 7.89 m/s2 for 10 seconds. How far has it traveled?

x = x0 + v0t + 1/2 at2

A car accelerates from rest at 7.89 m/s2 for 10 seconds. How far has it traveled?

x = x0 + v0t + 1/2 at2

X = 0 + (0 m/s)(10 s) + (1/2)(7.89 m/s2 )(10 s)2

x = 394.5 m

Free Fall

When an object is released, it falls towards earth due to the gravitational attraction of the Earth.

As objects fall, they will accelerate at a constant rate of 9.8 m/s2 regardless of mass.

***We will neglect air resistance at this time***

g = force of gravityor acceleration due to gravity

g = 9.8 m/s2

Velocity at a particular point of an objects fall is called instantaneous velocity.

v = vo + gt

initial velocity 9.8 m/s2 time elapsed

g = 9.8 m/s2

Velocity at a particular point of an objects fall is called instantaneous velocity.

v = vo + gt

initial velocity 9.8 m/s2 time elapsed

If the initial velocity is starting from rest, then vo = 0

And we can set up our equation as

v = gt

Example Problems

• A ball is dropped off of the edge of a building and it takes 2.3 seconds to reach the ground. What is the objects speed right before it hits the ground?

Example Problems

• A ball is dropped off of the edge of a building and it takes 2.3 seconds to reach the ground. What is the objects speed right before it hits the ground?

v = gt

Example Problems

• A ball is dropped off of the edge of a building and it takes 2.3 seconds to reach the ground. What is the objects speed right before it hits the ground?

v = gt

v = (9.8 m/s2)(2.3s)

v = 22.54 m/s

If time is not given, then we need to know the distance from which an object was dropped. In that case we will use the

following equation

vf2 = vo

2 + 2gd

(Velocity final)2 =(initial velocity)2 + 2(9.8 m/s2)(change in distance)

You will use this equation for

Workbook Pages # 10 – 11 Problems 9-14

Example: A ball is dropped off of the edge of a 25 m building. You are at a window 10 m above the ground. How fast is the ball travelling

when it passes your window?

vf2 = vo

2 + 2gd

A ball is dropped off of the edge of a 25 m building. You are at a window 10 m above the ground. How fast is the ball travelling

when it passes your window?

vf2 = vo

2 + 2gd

vf2 = 02 + 2(9.8m/s2)(25m - 10m)

A ball is dropped off of the edge of a 25 m building. You are at a window 10 m above the ground. How fast is the ball travelling

when it passes your window?

vf2 = vo

2 + 2gd

vf2 = 02 + 2(9.8m/s2)(25m - 10m)

√ vf2 = √(294)

vf = 17.146

“Falling down”

• Velocity is positive

• Change in position is +

“Up”

• Velocity is negative

• Change in position is negative

For Falling Objects

Reminder: Distance (x) is still calculated using our earlier equation

x = x0 + v0t + 1/2 at2

Where acceleration (a) is the acceleration due to gravity (g) or 9.8 m/s2

Example: You decide to figure out how tall a building is with out actually measuring it….. You drop a book

from the edge of the building and observe that it takes 3 s to reach the ground. How tall is the

building?

x = x0 + v0t + 1/2 at2

Example: You decide to figure out how tall a building is with out actually measuring it….. You drop a book

from the edge of the building and observe that it takes 3 s to reach the ground. How tall is the

building?

x = x0 + v0t + 1/2 at2

x = 0 + 0(3s) + 1/2 (9.8 m/s2)(3)2

x = 44.1 m

Example: How long is a ball up in the air when it is thrown

straight up with a speed of 0.75 m/s?

v = v0 + at

Rearrange to solve for t

And use g for a

How long is a ball up in the air when it is thrown straight up with a speed of 0.75 m/s?

v = v0 + at

Rearrange to solve for t

And use g for a

t = (Vf – Vi) = 0 - (-0.75) = 0.076 s

g 9.8 m/s2

Since this is only the time to reach the top of the path, we will need to double this time to find the time for the round trip.

0.076(2) = .152 seconds

Example: How far will a freely falling object have fallen from a position of rest

when it reaches a speed of 15 m/s?

• Two problem solving options!!!

vf2 = vo

2 + 2gd or v = v0 + at d = ½ at2

Use 10 m/s2 for g

Example: How far will a freely falling object have fallen from a position of rest

when it reaches a speed of 15 m/s?

• Two problem solving options!!!

vf2 = vo

2 + 2gd or v = v0 + at d = ½ at2

Use 10 m/s2 for g

Both equations should give you 11.25 as your answer!!!!!

The End!!!