Phys151_PPL4

Phys151_PPL4 - Describing U-A Motion Kinematic Equations a...

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Unformatted text preview: Describing U-A Motion Kinematic Equations a = constant v = v + a.t d = v .t + 1/2.a.t 2 Take v = 0, i.e. dropping from rest. Plots are as shown for a, v (velocity) and d vs. t (not to scale, positive is down) Note that d is really DISPLACEMENT! 10 20 30 40 50 60 70 80 90 t Series2 Series3 Series4 a = constant v = v + a.t d = v .t + 1/2.a.t 2 Describing U-A Motion Kinematic Equations a = constant v = v + a.t d = v .t + 1/2.a.t 2 Take v = +10 , i.e. thrown down. Plots are as shown for a, v and d vs t (not to scale, positive is down) 20 40 60 80 100 120 140 t Series2 Series3 Series4 a = constant v = v + a.t d = v .t + 1/2.a.t 2 Describing U-A Motion Kinematic Equations a = constant v = v + a.t d = v .t + 1/2.a.t 2 Take v = -10 , i.e. thrown up. Plots are as shown for a, v(elocity) and d(isplacement) vs t (not to scale, positive is down). The funny curve for the speed tells you that the kinematic equations are vector equations!-20-10 10 20 30 40 50 t Series2 Series3 Series4 a = constant v = v + a.t d = v .t + 1/2.a.t 2 Problem-solving I throw a ball up with a launch speed of 5 m/s. When does it reach its topmost point? Note: at its topmost point, the ball must be momentarily at rest (i.e. it has zero speed there). Strategy to be followed: 1. construct a diagram 2. identify and list given information, with appropriate signs 3. identify and list unknown information, with appropriate signs 4. identify equation to be used 5. substitute and do required algebra 6. check answer KEqs. a = constant v = v + a.t d = v .t + 1/2.a.t 2 v2 2 = v 2 + 2.a.d (2) and (3): v = - 5 m/s, v = 0, a = +10 m/s/s, t = ? (4) v = v + a.t (5) 0 = -5 + 10.t so that t = 0.5 s. (6) An upward moving object always loses 10 m/s in every second of travel. So answer is reasonable! Throwing the ball up!...
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Phys151_PPL4 - Describing U-A Motion Kinematic Equations a...

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