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g. Applications of the Kinematic Equations. h. v o. - H. AP Physics B Lecture Notes. v f = 0. a = g. h = D x. g. One Dimensional Motion. H = -50 m. v o =30 m/s. What is the maximum height the ball reaches?. g = -9.8 m/s 2. h. v 2 f = v 2 o + 2a D x. v o.

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  1. g Applications of the Kinematic Equations h vo -H AP Physics B Lecture Notes

  2. vf = 0 a = g h = Dx g One Dimensional Motion H = -50 m vo =30 m/s What is the maximum height the ball reaches? g = -9.8 m/s2 h v2f = v2o + 2a Dx vo 0 = v2o + 2(g)h -h =v2o/2g H = 45.9 m 2

  3. vf = 0 a = g t = tr g One Dimensional Motion H = -50 m vo =30 m/s How much time to reach the maximum height? g = -9.8 m/s2 h vf = vo + at vo 0 = vo + (g)tr H = 3.06 s 3

  4. a = g Dx = 0 g One Dimensional Motion H = -50 m vo =30 m/s What is the speed of the ball at xo = 0? g = -9.8 m/s2 h v2f = v2o + 2a Dx vo v2f = v2o + 2a (0) v2f = v2o H vf= -30.0 m/s 4

  5. vf = -30 m/s a = g t = t1 g One Dimensional Motion H = -50 m vo =30 m/s How much time to reach the xo = 0? g = -9.8 m/s2 h vo vf = vo + at vf = vo + at1 -30 m/s = 30 m/s + (-9.8 m/s2)t1 H = 6.12 s 5

  6. a = g Dx = H g One Dimensional Motion H = -50 m vo =30 m/s What is the final speed of the ball? g = -9.8 m/s2 h v2f = v2o + 2a Dx vo v2f = v2o + 2a (-50m) v2f = (30 m/s)2 + 2(-9.8 m/s2(-50m) H vf = -43.36 m/s 6

  7. vf = -43.36 m/s a = g t = tT g One Dimensional Motion H = -50 m vo =30 m/s How much total time in the air? g = -9.8 m/s2 h vo vf = vo + at vf = vo + atT -43.36 m/s = 30 m/s + (-9.8 m/s2)tT H = 7.49 s -vf 7

  8. One Dimensional Motion 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? vf = 0 = 90 m 8

  9. 11 88 m 12 m t t2 t1 One Dimensional Motion 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? 9

  10. 11 88 m 12 m t t2 t1 0 10

  11. 0 0 One Dimensional Motion On a dry road a car with good tires is able to brake with a constant deceleration of 4.92 m/s2. 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? 11

  12. (m/s) vmax (s) One Dimensional Motion The maximum acceleration of a subway train is 1.34 m/s2, and subway stations are located 806 m apart, (a) what is the maximum speed a subway train can attain between stations? a = 1.34 m/s2 403 m 403 m t 12

  13. (m/s) a = 1.34 m/s2 vmax x1 = 403 m x2 = 403 m (s) t 0 13

  14. (m/s) a = 1.34 m/s2 vmax x1 = 403 m x2 = 403 m t1 t2 (s) t 0 (b) what is the travel time between stations? 14

  15. One Dimensional Motion END

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