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/s 2 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 = 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 Motion Through derivations (a fancy way of saying use algebra to rearrange your problem ) we can rewrite a =  v/t in two new ways: v = v 0 + at x = x 0 + v 0 t + 1/2 at 2 V 0 is the initial velocity X is the final distance x 0 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/s 2 for 3 seconds. What is the skateboarder’s velocity after 4 seconds? v = v 0 + at

Example Problem A skateboarder traveling at 1 m/s accelerates at 1.5 m/s 2 for 3 seconds. What is the skateboarder’s velocity after 4 seconds? v = v 0 + at v = 1 m/s + (1.5 m/s 2 )(3 s) v = 5.5 m/s

Journal Practice A car traveling at 5 m/s accelerates at 3 m/s 2 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/s 2 for 8 seconds. What is the car’s velocity after 8 seconds? v = v 0 + at

Journal Practice A car traveling at 5 m/s accelerates at 3 m/s 2 for 8 seconds. What is the car’s velocity after 8 seconds? v = v 0 + at v = (5 m/s) + (3 m/s 2 )(8) = 29 m/s

Example A car traveling at 8 m/s accelerates at 5 m/s 2 for 10 seconds. How far has the car traveled 10 seconds after it began accelerating? x = x 0 + v 0 t + 1/2 at 2

Example A car traveling at 8 m/s accelerates at 5 m/s 2 for 10 seconds. How far has the car traveled 10 seconds after it began accelerating? x = x 0 + v 0 t + 1/2 at 2 x = x 0 + (8 m/s) (10 s) + 1/2 (5 m/s 2 )(10 s) 2 Set initial distance or starting point as 0. x = 0 m + (8 m/s) (10 s) + 1/2 (5 m/s 2 )(10 s) 2 X= 330 m

Journal Practice A car traveling at 5 m/s accelerates at 3 m/s 2 for 8 seconds. How far has the car traveled 2 seconds after it began accelerating? x = x 0 + v 0 t + 1/2 at 2

Journal Practice A car traveling at 5 m/s accelerates at 3 m/s 2 for 8 seconds. How far has the car traveled 2 seconds after it began accelerating? x = x 0 + v 0 t + 1/2 at 2 t = 2 seconds x 0 = 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/s 2 for 8 seconds. How far has the car traveled 2 seconds after it began accelerating? x = x 0 + v 0 t + 1/2 at 2 X = 0 + (5 m/s)(2 s) + (1/2)(3 m/s 2 )(2 s) 2 X = 16 m

Journal A car accelerates from rest at 7.89 m/s 2 for 10 seconds. How far has it traveled? x = x 0 + v 0 t + 1/2 at 2

A car accelerates from rest at 7.89 m/s 2 for 10 seconds. How far has it traveled? x = x 0 + v 0 t + 1/2 at 2 X = 0 + (0 m/s)(10 s) + (1/2)(7.89 m/s 2 )(10 s) 2 x = 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/s 2 regardless of mass. ***We will neglect air resistance at this time***

g = force of gravity or acceleration due to gravity g = 9.8 m/s 2 Velocity at a particular point of an objects fall is called instantaneous velocity. v = v o + gt initial velocity 9.8 m/s 2 time elapsed

g = 9.8 m/s 2 Velocity at a particular point of an objects fall is called instantaneous velocity. v = v o + gt initial velocity 9.8 m/s 2 time elapsed If the initial velocity is starting from rest, then v o = 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/s 2 )(2.3s) v = 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 v f 2 = v o 2 + 2g  d (Velocity final) 2 =(initial velocity) 2 + 2(9.8 m/s 2 )(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? v f 2 = v o 2 + 2g  d

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? v f 2 = v o 2 + 2g  d v f 2 = (9.8m/s 2 )(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? v f 2 = v o 2 + 2g  d v f 2 = (9.8m/s 2 )(25m - 10m) √ v f 2 = √(294) v f =

“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 = x 0 + v 0 t + 1/2 at 2 Where acceleration (a) is the acceleration due to gravity (g) or 9.8 m/s 2

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 = x 0 + v 0 t + 1/2 at 2

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 = x 0 + v 0 t + 1/2 at 2 x = 0 + 0(3s) + 1/2 (9.8 m/s 2 )(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 = v 0 + 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 = v 0 + at Rearrange to solve for t And use g for a t = (V f – V i ) = 0 - (-0.75) = s g 9.8 m/s 2 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 (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!!! v f 2 = v o 2 + 2g  d or v = v 0 + at d = ½ at 2 Use 10 m/s 2 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!!! v f 2 = v o 2 + 2g  d or v = v 0 + at d = ½ at 2 Use 10 m/s 2 for g Both equations should give you as your answer!!!!!

The End!!!