Presentation on theme: "One Dimensional Motion"— Presentation transcript:
1 One Dimensional Motion Pre AP PhysicsOne Dimensional MotionVelocity, Acceleration, Motion Graphs
2 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 quantitySpeed is scalar and is distance divided by time.
3 Types of VelocityAverage 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, Vi = zero.Final Velocity – the velocity of the object at the end of the time interval. For an object braking to a stop, Vf = zero.Terminal Velocity – got to wait for the force chapter, sorry.
8 ClassWorkHow far have you walked if your velocity is 2m/s West and you walk for 2 minutes?How much time would it take to drive 2km if your average velocity is 30m/s?What is Mr. Crabtree’s average velocity if he runs 3.2km in a time of 18 minutes and 20 seconds?
9 Distance – Time Graphs Also called Position – Time Graphs Position (m)Time (sec)
11 An unchanging Velocity Constant VelocityA special CaseAn unchanging VelocityNot speeding up, not slowing down, not changing direction.Its like perfect cruise control on a level, straight highway.
12 Change in Velocity speed up, slow down, or change in direction AccelerationChange in Velocityspeed up, slow down, or change in directionUnits= m/ssm/s ms-2ā= V2-V1t2-t1
13 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
14 For an object accelerating @ 3m/s2 An acceleration value tells you by how much an object is changing its velocityFor an object 3m/s2Time Velocity0 sec O m/s1 sec m/s2 sec m/s3 sec m/s4 sec m/s5 sec m/s
15 Example ProblemThe 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 SolutionVi= 2m/s a= v a= 16m/s – 2 m/sVf= 16m/s t sec- 1 secT1= 1 sec a= 14m/sT2= 4.5 sec sec= 4m/s2
16 Look at the followingPractice Problemsp.491,2,3p.49SampleOn A velocity-TimeGraph:Slope= Acceleration
17 Sample 2B p.49A shuttle bus slows to a stop with an average acceleration of –1.8m/s2. How long does it take the bus to slow from 9m/s to 0m/s?
18 PracticeWhat 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
19 #1. Given Formula Solution Vi= 4m/s a = Vf-ViVf= 36m/s tt= 4s – 44#2. Vi= 36 m/s a = Vf-ViVf= 15 m/s tt= 3 sec – 3638 m/s2- 7 m/s2
20 #3. Vi= -3 m/s a= Vf – Vi Vf= + 4.5 m/s t t= 2.5 sec 4.5 – (-3) 2.5 GREAT JOB!
21 Vi= Vf – at t = Vf – Vi a= Vf – Vi a t Acceleration that does not change is called uniform or constant acceleration(for simplicity sake, the problems we will solve will be in uniform acceleration)2nd Formula–Vf= Vi + atVi= Vf – at t = Vf – Vi a= Vf – Via t
22 Given Formula Solution Vi= 2 m/s Vf= Vi + at A = 4 m/s2 Vf= 2 + 4(2.5) If a car with a velocity of 2 m/s accelerates at a rate of (+)4 m/s2 for 2.5 sec., what is the velocity at 2.5 sec?Given Formula SolutionVi= 2 m/s Vf= Vi + atA = 4 m/s Vf= 2 + 4(2.5)T= 2.5 secVf= ??Vf = 12 m/sDo sample problem p. 55
23 Sample Problem p.55A plane starting from rest at one end of the runway undergoes a uniform acceleration of 4.8m/s2 for 15 seconds before takeoff. What is the speed at takeoff?
24 Displacement when velocity and time are known d= ½ (Vf + Vi) t *True when object is accelerated uniformlyDisplacement when acceleration and time are knownd= Vit + ½ at2*recall that an object starting from rest has Vi= 0 m/sWhen starting from restd= ½ at2
25 Sample ProblemA plane starting from rest at one end of the runway undergoes a uniform acceleration of 4.8m/s2 for 15 seconds before takeoff. How long must the runway be for the plane to be able to take off?
27 Page 55 practice problems 1-4 2 DPage 55 practice problems 1-4
28 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.
29 Vf= Vi2+ 2ad a= Vf2 – Vi2 d= Vf2 – Vi2 2d 2a Solving without timeVf2= Vi2 + 2 adVf= Vi2+ 2ad a= Vf2 – Vi d= Vf2 – Vi22d aGiven Formula SolutionVi= a= Vf2- Vi a =Vf= 71 m/s d (1000)d= 1000m2.5 m/s2HW pages
36 Acceleration due to gravity Earth’s gravity pulls everything towards the center of the earthAll 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 accelerationGravity 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
37 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 gravitysize and mass determine gravityacceleration due to gravity is a vector quantity
38 We will use an average value for gravity when solving problems on Earth. g= m/s2As an object falls w/out air on EarthTime Velocity0s1s m/s ft/s2s m/s ft/s3s m/s ft/s4s m/s ft/s
39 Displacement During free fall Displacement = ½ a t2 for objects falling from rest. a=accel due to gravity, -9.8m/s2.After falling for 1 second, an object on Earth has fallen -4.9mAfter 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.
40 Practice problem p.68U-V 3-7 **overhead model**as the ball goes up gravity takes 9.8 m/s off the velocity for every secondas the ball goes down gravity puts on 9.8 m/s for every second the ball fallsExample problems p.68UPractice problem p.68U-V 3-7reaction timer page 68T
41 Gravity is constant and therefore we can “sub” it into ALL of our formulas g = vt2. Vf = Vi + gt4. D = Vit+ ½ gt25. Vf2 = Vi2 + 2 gd
42 *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?Given Formula Solutiong= m/s Vf = Vi + gt Vf= 0 + (-9.8)(1.5)Vi= 0 m/s d= Vit + ½ gt2t = 1.5 s d= 0 + ½ (-9.8)(1.5)2(A): Vf= - 15 m/s(B): d= -11 m
43 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 Solutiong= m/s Vf= Vi + gt Vf= 0 + (-9.8)4t= 4 sec d= Vit + ½ gt (A): Vf= -39.2m/sVi= 0 m/s d= ½ (-9.8)(4)2( B): d= m
44 Formulas a= Vf – Vi Vi Vf a t t Vf= Vi+ at Vi Vf a t d= ½ (Vf + Vi)t Vi Vf d td= Vit + ½ at Vi d a tVf2= Vi2 +2ad Vi Vf d ag= Vf – Vi Vi Vf g tVf= Vi + gt Vi Vf g td= Vit + ½ gt Vi d g tVf2= Vi2 + 2gd Vi Vf d g