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Vovo g h HH Applications of the Kinematic Equations AP Physics B Lecture Notes.

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Presentation on theme: "Vovo g h HH Applications of the Kinematic Equations AP Physics B Lecture Notes."— Presentation transcript:

1 vovo g h HH Applications of the Kinematic Equations AP Physics B Lecture Notes

2 vovo g h  H H = -50 m v o =30 m/s g = -9.8 m/s 2 One Dimensional Motion What is the maximum height the ball reaches? v 2 f = v 2 o + 2a  x 0 = v 2 o + 2(g)h -h =v 2 o /2g v f = 0 a = g h =  x = 45.9 m 2

3 vovo g h  H H = -50 m v o =30 m/s g = -9.8 m/s 2 How much time to reach the maximum height? v f = v o + at 0 = v o + (g)t r v f = 0 a = g t = t r = 3.06 s One Dimensional Motion 3

4 vovo g h  H H = -50 m v o =30 m/s g = -9.8 m/s 2 What is the speed of the ball at x o = 0? a = g  x = 0 v f = m/s v 2 f = v 2 o + 2a  x v 2 f = v 2 o + 2a  v 2 f = v 2 o One Dimensional Motion 4

5 vovo g h  H H = -50 m v o =30 m/s g = -9.8 m/s 2 How much time to reach the x o = 0? v f = v o + at v f = -30 m/s a = g t = t 1 = 6.12 s v f = v o + at m/s = 30 m/s + (-9.8 m/s 2 )t 1 One Dimensional Motion 5

6 vovo g h  H H = -50 m v o =30 m/s g = -9.8 m/s 2 What is the final speed of the ball? a = g  x = H = m/s v 2 f = v 2 o + 2a  x v 2 f = v 2 o + 2a  m  vfvf v 2 f = (30 m/s) 2 + 2(-9.8 m/s 2  m  One Dimensional Motion 6

7 vovo g h  H H = -50 m v o =30 m/s g = -9.8 m/s 2 How much total time in the air? v f = v o + at v f = m/s a = g t = t T = 7.49 s v f = v o + at T m/s = 30 m/s + (-9.8 m/s 2 )t T vfvf One Dimensional Motion 7

8 A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.0 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops? v f = 0 = 90 m One Dimensional Motion 8

9 A sprinter has a top speed of 11.0 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 12.0 m. He is then able to maintain this top speed for the remainder of a 100 m race. What is his time for the 100 m race? t t1t1 12 m t2t2 88 m 11 One Dimensional Motion 9

10 11 t t1t1 t2t2 12 m 88 m 0 10

11 On a dry road a car with good tires is able to brake with a constant deceleration of 4.92 m/s 2. If the car is initially traveling at 24.6 m/s, (a) how much time is required to stop? (b) how far does it travel in this time? 0 0 One Dimensional Motion 11

12 The maximum acceleration of a subway train is 1.34 m/s 2, and subway stations are located 806 m apart, (a) what is the maximum speed a subway train can attain between stations? v max t a =  1.34 m/s m (m/s) (s) One Dimensional Motion 12

13 v max t a =  1.34 m/s 2 x 1 = 403 m (m/s) (s) x 2 = 403 m 0 13

14 0 v max t a =  1.34 m/s 2 x 1 = 403 m (m/s) (s) x 2 = 403 m t1t1 t2t2 (b) what is the travel time between stations? 14

15 END One Dimensional Motion


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