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Unformatted text preview: Describing UA 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 UA 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 UA 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!2010 10 20 30 40 50 t Series2 Series3 Series4 a = constant v = v + a.t d = v .t + 1/2.a.t 2 Problemsolving • 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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 Spring '09
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 Force, Mass, net force, Law of Force

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