Presentation on theme: "Acceleration - rate of change in velocity per unit time."— Presentation transcript:
1Acceleration - rate of change in velocity per unit time.
2Acceleration = ∆v / ∆t (vf - vi / tf - ti) 8m/s - 4m/s = 4m/s = 2m/s24s - 2s sAcceleration’s units are distance per time squaredWhat is the graph of an object that is decelerating?What is the graph of an object at constant acceleration?
3Remember 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?
4The 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).eastABvelocity(m/s)CDtime (s)EwestThe sign of acceleration does not indicate whether or notThe object is speeding up or slowing down (only its direction).
5A – 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).
6Final Velocity with Constant Acceleration a = ∆v OR ∆v = a ∆t OR vf - vi = a ∆t∆tvf = vi + a ∆tExample 1: A bus that is traveling at 30.0 km/hr speeds up at aconstant rate of 3.5 m/s2. What velocity does itreach 6.8 s later?Example 2: A car slows from 22 m/s to 3.0 m/s at a constant rateof 2.1 m/s2. How many seconds are required beforethe car is traveling at 3.0 m/s?
7Position 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?
8Free 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/s2Sun: g = 275 m/s2Venus: g = 8.9 m/s2Moon: g = 1.6 m/s2Jupiter: g = 25 m/s2
9This 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 + atfdf = di + vitf + 1/2at2vf2 = vi2 + 2a(df - di)
10ExamplesYou 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?