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Pre AP Physics One Dimensional Motion Velocity, Acceleration, Motion Graphs

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Velocity ● Displacement divided by time interval ● Velocity is the change in position divided by the time interval during which the motion took place. ● Velocity is a vector quantity ● Speed is scalar and is distance divided by time.

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Types of Velocity ● Average Velocity – the total displacement of motion divided by the total time interval. ● Instantaneous Velocity – the velocity in a specific instant. Like a quick check on your speedometer. ● Initial Velocity – the velocity at the start of the time interval. When object starts at rest, V i = zero. ● Final Velocity – the velocity of the object at the end of the time interval. For an object braking to a stop, V f = zero. ● Terminal Velocity – got to wait for the force chapter, sorry.

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Average Velocity Equation Vavg=

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Manipulate to solve for d and t

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Velocity and Direction

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Practice Problems 2A p.44 1-5

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ClassWork 1. How far have you walked if your velocity is 2m/s West and you walk for 2 minutes? 2. How much time would it take to drive 2km if your average velocity is 30m/s? 3. What is Mr. Crabtree’s average velocity if he runs 3.2km in a time of 18 minutes and 20 seconds?

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Distance – Time Graphs Also called Position – Time Graphs Position (m) Time (sec)

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Slope – what does it mean?

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Constant Velocity ● A special Case ● An unchanging Velocity ● Not speeding up, not slowing down, not changing direction. ● Its like perfect cruise control on a level, straight highway.

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Change in Velocity speed up, slow down, or change in direction ā= V 2 -V 1 t 2 -t 1 Units = m/s s m/s 2 ms -2

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Acceleration is a vector quantity (like velocity) Direction makes a difference! If you have (+) positive velocity and speed up =(+)positive acceleration (+) positive velocity and slow down =(-)negative acceleration (+) positive velocity and slow down =(-)negative acceleration

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An acceleration value tells you by how much an object is changing its velocity Time Velocity 0 sec O m/s 1 sec 3 m/s 2 sec 6 m/s 3 sec 9 m/s 4 sec 12 m/s 5 sec 15 m/s For an object accelerating @ 3m/s 2

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Example Problem The velocity of a car increases from 2 m/s at 1 sec. To 16 m/s at 4.5 sec. What is the car’s average acceleration? Given Formula Solution V i = 2m/s a= v a= 16m/s – 2 m/s V f = 16m/s t 4.5 sec- 1 sec T 1 = 1 sec a= 14m/s T 2 = 4.5 sec 3.5 sec 4m/s 2 = 4m/s 2

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Look at the following Practice Problems p.491,2,3 p.49Sample On A velocity-Time Graph: Slope= Acceleration

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Sample 2B p.49 ● A shuttle bus slows to a stop with an average acceleration of –1.8m/s 2. How long does it take the bus to slow from 9m/s to 0m/s?

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Practice ● What is the acceleration of a car that goes from 4m/s to 36m/s in a time interval of 4 seconds? ● What is the acceleration of a car that goes from 36m/s to 15m/s in a time interval of 3 seconds? ● What is the acceleration of a vehicle that goes from –3m/s to 4.5m/s in a time of 2.5 seconds? ● Solutions on the next 2 slides

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#1. Given Formula Solution V i = 4m/s a = V f -V i V i = 4m/s a = V f -V i V f = 36m/s t V f = 36m/s t t = 4s 36 – 4 t = 4s 36 – 4 4 #2. V i = 36 m/s a = V f - V i V f = 15 m/s t V f = 15 m/s t t = 3 sec 15 – 36 t = 3 sec 15 – 36 3 8 m/s 2 - 7 m/s 2

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#3. V i = -3 m/s a= V f – V i V f = + 4.5 m/s t V f = + 4.5 m/s t t= 2.5 sec 4.5 – (-3) t= 2.5 sec 4.5 – (-3) 2.5 2.5 3 m/s 2 GREAT JOB!

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Acceleration that does not change is called uniform or constant acceleration (for simplicity sake, the problems we will solve will be in uniform acceleration) 2 nd Formula– V f = V i + at V f = V i + at V i = V f – at t = V f – V i a= V f – V i V i = V f – at t = V f – V i a= V f – V i a t a t

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● If a car with a velocity of 2 m/s accelerates at a rate of (+)4 m/s 2 for 2.5 sec., what is the velocity at 2.5 sec? Given Formula Solution V i = 2 m/s V f = V i + at A = 4 m/s 2 V f = 2 + 4(2.5) T= 2.5 sec V f = ?? V f = 12 m/s Do sample problem p. 55

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Sample Problem p.55 ● A plane starting from rest at one end of the runway undergoes a uniform acceleration of 4.8m/s 2 for 15 seconds before takeoff. What is the speed at takeoff?

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Displacement when velocity and time are known Displacement when velocity and time are known d= ½ (V f + V i ) t d= ½ (V f + V i ) t * True when object is accelerated uniformly Displacement when acceleration and time are known d= V i t + ½ at 2 * recall that an object starting from rest has V i = 0 m/s When starting from rest d= ½ at 2 d= ½ at 2

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Sample Problem ● A plane starting from rest at one end of the runway undergoes a uniform acceleration of 4.8m/s 2 for 15 seconds before takeoff. How long must the runway be for the plane to be able to take off?

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2 C Page 53 practice problems 1-4

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2 D Page 55 practice problems 1-4

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● A certain airplane wants to takeoff from a runway that is 1 kilometer long. The plane starts from rest and must get to a speed of 71m/s in order to lift off the ground. What is the minimum acceleration the pilot has to achieve to order to takeoff? ** you are not given time. ● Answer is on the next slide.

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Solving without time V f 2 = V i 2 + 2 ad V f = V i 2 + 2ad a= V f 2 – V i 2 d= V f 2 – V i 2 2d 2a 2d 2a Given Formula Solution V i = 0 a= V f 2 - V i 2 a = 71 2 - 0 V f = 71 m/s 2d 2(1000) d= 1000m 2.5 m/s 2 HW pages

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2 E Page 58 practice problems 1-4

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Velocity Time Graphs Velocity (m/s) Time (sec)

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V – t Graphs ● Slope equals ______________________ ● Area under the slope (curve, line) equals ________________

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examples Velocity (m/s) Time (sec)

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Class Work Page 68 N and P

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D-t and V-t graphs side by side.

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Acceleration due to gravity Earth’s gravity pulls everything towards the center of the earth All things are pulled at the same rate. (atoms to elephants- all the same) – so long as we neglect air resistance, which we will most of the time. Gravity is a uniform acceleration Gravity on Earth can change as you go from place to place. Gravity changes as you change the distance from the center of the earth to your location. EXAMPLE: on a mountain gravity is less than at sea level. also earth isn’t perfectly round gravity is less at equator than poles

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Gravity can vary from place to place; but only at the same place are all objects accelerated the same. Gravity can vary from place to place; but only at the same place are all objects accelerated the same. Every planet- (as well as the moon and sun) has its own acceleration due to gravity Every planet- (as well as the moon and sun) has its own acceleration due to gravity size and mass determine gravity size and mass determine gravity acceleration due to gravity is a vector quantity acceleration due to gravity is a vector quantity

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● We will use an average value for gravity when solving problems on Earth. g= - 9.81 m/s 2 ● As an object falls w/out air on Earth Time Velocity Time Velocity 0s 0 0s 0 1s -9.8 m/s -32 ft/s 1s -9.8 m/s -32 ft/s 2s -19.6 m/s -64 ft/s 2s -19.6 m/s -64 ft/s 3s -29.4 m/s -96 ft/s 3s -29.4 m/s -96 ft/s 4s -39.2 m/s -128 ft/s 4s -39.2 m/s -128 ft/s

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Displacement During free fall ● Displacement = ½ a t 2 for objects falling from rest. a=accel due to gravity, -9.8m/s 2. ● After falling for 1 second, an object on Earth has fallen -4.9m ● After falling for 2 seconds, the object is -19.6m below the drop position. ● After falling for 3 seconds, the object is -44.1m below the drop position.

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**overhead model** ● as the ball goes up gravity takes 9.8 m/s off the velocity for every second ● as the ball goes down gravity puts on 9.8 m/s for every second the ball falls ● Example problems p.68U ● Practice problem p.68U-V 3-7 ● reaction timer page 68T

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● Gravity is constant and therefore we can “sub” it into ALL of our formulas 1. g = v t 2. V f = V i + gt 4. D = V i t+ ½ gt 2 5. V f 2 = V i 2 + 2 gd

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*Class work* The time the sky screamer ride at Astro world is free falling is 1.5 sec, (A) what is its velocity at the end of the time? (B) How far does it fall? ● The time the sky screamer ride at Astro world is free falling is 1.5 sec, (A) what is its velocity at the end of the time? (B) How far does it fall? Given Formula Solution g = - 9.8 m/s 2 V f = V i + g t V f = 0 + ( - 9.8)(1.5) V i = 0 m/s d= V i t + ½ g t 2 t = 1. 5 s d= 0 + ½ ( - 9.8)(1.5) 2 (A): V f = - 15 m/s (B): d= -11 m

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● A brick falls freely from a high scaffold (A) what is the velocity after 4 seconds (B) how far does the brick fall during these 4 seconds? Given Formula Solution g= - 9.8 m/s 2 V f = V i + gt V f = 0 + (-9.8)4 t= 4 sec d= V i t + ½ gt 2 (A): V f = -39.2m/s V i = 0 m/s d= ½ (-9.8)(4) 2 ( B): d= - 78.4m ( B): d= - 78.4m

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Formulas a= V f – V i V i V f a t t V f = V i + at V i V f a t d= ½ (V f + V i )t V i V f d t d= V i t + ½ at 2 V i d a t V f 2 = V i 2 +2ad V i V f d a g= V f – V i V i V f g t t V f = V i + gt V i V f g t d= V i t + ½ gt 2 V i d g t V f 2 = V i 2 + 2gd V i V f d g

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What’s going on here …..?

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