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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?

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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?

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Physics More Linear Motion!!!!

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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!

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When reading a word problem identify what you are given –These are your “knowns” Identify what you are solving for –This is your “unknown”

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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

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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

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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

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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?

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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

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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

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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

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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

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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

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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!!!

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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

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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

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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

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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***

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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

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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

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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?

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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

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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

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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

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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

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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)

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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 =

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“Falling down” Velocity is positive Change in position is + “Up” Velocity is negative Change in position is negative

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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

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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

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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

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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

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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

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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

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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!!!!!

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The End!!!

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