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Acceleration - rate of change in velocity per unit time.

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Presentation on theme: "Acceleration - rate of change in velocity per unit time."— Presentation transcript:

1 Acceleration - rate of change in velocity per unit time.

2 Acceleration = ∆v / ∆t (vf - vi / tf - ti)
8m/s - 4m/s = 4m/s = 2m/s2 4s - 2s s Acceleration’s units are distance per time squared What is the graph of an object that is decelerating? What is the graph of an object at constant acceleration?

3 Remember a distance vs. time graph of an object that is accelerating is quadratic.
The rate of velocity is not constant. Since the velocity is changing per unit time so to is the distance traveled. What is the distance-time graph of an object that is decelerating?

4 The sign of acceleration does not indicate whether or not
Using the graph explain the acceleration and position of these 5 runners (assume east is positive). east A B v e l o c i t y (m/s) C D time (s) E west The sign of acceleration does not indicate whether or not The object is speeding up or slowing down (only its direction).

5 A – Is traveling at a constant velocity (acceleration = 0) towards the east.
B – Is traveling towards the east, but its velocity is increasing (constant acceleration). C – Is heading in the east direction, but slowing down (constant deceleration). D – Is heading in the west direction but is slowing down (constant deceleration) until it stops, then heads to the east and starts increasing its velocity (constant acceleration). E – Is heading west at a constant velocity (acceleration = 0).

6 Final Velocity with Constant Acceleration
a = ∆v OR ∆v = a ∆t OR vf - vi = a ∆t ∆t vf = vi + a ∆t Example 1: A bus that is traveling at 30.0 km/hr speeds up at a constant rate of 3.5 m/s2. What velocity does it reach 6.8 s later? Example 2: A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is traveling at 3.0 m/s?

7 Position with Constant Acceleration df = di + vitf + 1/2atf2 Velocity with Constant Acceleration vf2 = vi2 + 2a(df - di) Example 1: Sekazi is learning to ride a bike. His father pushes him with a constant acceleration of 0.50m/s2 for 6.0s, and then Sekazi continues at 3.0m/s for 6.0s before falling. What is Sekazi’s displacement? Example 2: Sunee is training for an upcoming 5.0km race. She starts out her training run by moving at a constant pace of 4.3m/s for 19min. Then she accelerates at a constant rate until she crosses the finish line 19.4s later. What is her acceleration during the last portion of the training run?

8 Free Fall - the motion of an object due to the force of gravity (air resistance = 0).
Neglecting air resistance, all objects in free fall have the same acceleration. Earth: g (acceleration due to gravity) = 9.8 m/s2 Sun: g = 275 m/s2 Venus: g = 8.9 m/s2 Moon: g = 1.6 m/s2 Jupiter: g = 25 m/s2

9 This same acceleration constant (9
This same acceleration constant (9.8 m/s2) applies to objects falling towards earth or moving away from earth. When solving for free fall these same equations apply: vf = vi + atf df = di + vitf + 1/2at2 vf2 = vi2 + 2a(df - di)

10 Examples You toss a ball up in the air at 35 m/s. What is the velocity of the ball after 2.2 s? 4.4 s? 7.0 s? A construction worker accidentally drops a brick for a high scaffold. What is the velocity of the brick after 4.0 s? How high was the scaffold? A tennis ball is thrown straight up with an initial speed of 22.5 m/s. How high does the ball rise? If the ball is caught at the same distance above the ground how long was the ball in the air? The current world record for vertical leap is 1.5 m. What was this individual’s initial speed? During this jump how long was the person in the air?


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