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Zoom.  Acceleration (a)– the rate of change of velocity (can be +, -, or centripetal)  Deceleration– only negative acceleration (slowing down)  Centripetal.

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Presentation on theme: "Zoom.  Acceleration (a)– the rate of change of velocity (can be +, -, or centripetal)  Deceleration– only negative acceleration (slowing down)  Centripetal."— Presentation transcript:

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2  Acceleration (a)– the rate of change of velocity (can be +, -, or centripetal)  Deceleration– only negative acceleration (slowing down)  Centripetal Acceleration– acceleration that occurs when changing direction (turning)  Freefall– objects falling due to gravity  Force is what you feel when you accelerate

3  Since acceleration is a change in velocity and velocity has 2 parts  We can accelerate by  Changes in speed 1. Speed up (+ acceleration) 2. Slow down (- acceleration)  Changing Direction 3. Turning  Changing both speed and direction

4  We don’t see acceleration of an object well with our eyes  But if we look at a speedometer we can see speed change.  Acceleration is a picture of what your speedometer is doing

5 SLOW ACCELERATIONFAST(ER) ACCELERATION

6 THE VARIABLESTHE GRAPH  Speed ( or velocity)is the dependent variable  Time is the independent variable Time Speed Graph setup = Formula a=a=  sp tt

7  Acceleration(  sp/  t)= slope on an acceleration graph (rise/run) Acceleration Graph Time Speed + acceleration (gas pedal) Constant Speed (0 acceleration) (cruise control) - Acceleration (deceleration) (brake pedal) No motion - stopped

8 Positive (+) slope = + accel (speeding up) Negative (-) slope = - accel (slowing down) Zero (0) slope (flat line) = 0 accel (constant speed) speed t t t

9 Steeper slopes show greater acceleration Straight lines = Constant acceleration Steady increase or decrease in speed speed t t Accelerates quickly Accelerates slowly decelerates slowly decelerates quickly When is the speed = to 0 (stopped)?

10  Describe which sport you think this graph represents and why.  Golf  Skydiving  Fishing  100-meter dash  Drag racing Speed Time

11  Rise–change in the graph on the vertical axis (Speed)  Run–change in the graph on the horizontal axis (Time) Acceleration= sp 2 -sp 1 t 2 -t 1 Slope = = Y 2 -y 1 X 2 -x 1 rise run P1 (0 s, 0 km/h) t 1 sp 1 P2 (20 s, 20 km/h) t 2 sp 2 P1P2 a=a= sp 2 -sp 1 t2-t1t2-t1 = 20 km/h 20 s = = 1 km/h/s 20 km/h - 0 km/h 20 s – 0 s

12  What is the acceleration of the object from 0 – 20 s?  What is the speed of the object from 20 – 40 s?  What is the speed of the object from 40 – 60 s? Acceleration(slope) = sp 2 – sp 1 (rise) t 2 – t 1 (run) = (20–0 m/s) ( 20 – 0 s) = 20 m/s 20 s = 1 m/s/s Acceleration(slope) = sp 2 – sp 1 (rise) t 2 – t 1 (run) = (20–20 m/s) ( 40 – 20 s) = 0 m/s 20 s = 0 m/s/s Acceleration(slope) = sp 2 – sp 1 (rise) t 2 – t 1 (run) = (0–20 m/s) ( 60 – 40 s) = -20 m/s 20 s = -1 m/s/s

13 Acceleration= final speed – initial speed final time – intial time m/s, cm/min, km/h s, min, hr m/s/s, cm/min/min, km/h/s Since: sp 2 = final speed, sp 1 = initial speed t 2 = final time, t 1 = initial time a = (sp f – sp i ) (t f – t i ) a = (sp 2 – sp 1) (t 2 – t 1 )

14 Acceleration= final speed – start speed final time – start time  or x a tt  sp ÷  sp = sp 2 – sp 1  t = t 2 – t 1 sp2=final speed, sp1=start speed t2=final time, t1=start time Since: Then: a =  sp  t

15  or x a tt  sp ÷ A kid is sledding down a hill. When he enters the course his speed is 10 m/s. After 2 seconds he has reached a speed of 26 m/s. What was his acceleration? GivenFormulaSetupSolution List Given Info And Unknown: sp 1 = 10 m/s sp 2 = 26 m/s  t = 2 s a= ? Show Needed Equations: a = Substitute with Units & Show Math: a = Circles Answer with Units: a = a = 8 m/s/s (sp 2 – sp 1 ) tt 26 m/s-10 m/s 2 s 16 m/s 2 s

16  Freefall: Objects falling due to gravity  All objects accelerate at the same rate due to gravity during freefall regardless of their mass  Objects may reach terminal velocity sooner due to air resistance  Gravitational acceleration (g) = 9.8 m/s/s on earth

17  or x a tt  sp ÷ A woman jumps from a plane. She opens her parachute five seconds later. How fast was she going when she opened her shoot (disregard air resistance)? GivenFormulaSetupSolution List Given Info And Unknown:  sp = ??  t = 5 s a= g=9.8 m/s/s Show Needed Equations:  sp = Substitute with Units & Show Math:  sp = Circles Answer with Units:  sp = 49.2 m/s a ·  t 9.8 m/s/s· 5 s


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