NOTES p.14. Deriving the Equations of Motion The following 3 equations can be used for any situation that involves a constant acceleration (horizontally.

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NOTES p.14

Deriving the Equations of Motion The following 3 equations can be used for any situation that involves a constant acceleration (horizontally OR vertically). ** MEMORIZE THE DERIVATIONS** GRAPH FOR CONSTANT ACCELERATION (in general) velocity (ms -1 ) time (s) u v t u = initial velocity (ms -1 ) v = final velocity (ms -1 ) a = acceleration (ms -2 ) s = displacement (m) t = time (s)

velocity (m/s) time (s) u v t 1 st Equation: “a” = rate of change of velocity a= (v - u) / t so v= u + at

velocity (ms -1 ) time (s) u v t 2 nd Equation: “s” = area under graph line Area A Area B s= Area A + Area B s = ut + ½ (v-u) t s= ut + ½ (at) t s= ut + ½ a t 2

v 2 = (u + at) 2 v 2 = u 2 + 2uat + (at) 2 v 2 = u 2 + 2uat + a 2 t 2 v 2 = u 2 + 2a (ut + ½ a t 2 ) v 2 = u 2 + 2as velocity (m/s) time (s) u v t 3 rd Equation: SQUARE EQUATION 1 Area A Area B

v 2 = u 2 + 2as s = ut + ½ a t 2 v = u + at For all calculations … 1 st Decide a SIGN CONVENTION 2 nd List “u v a s t” 3 rd Select equation(s) and calculate UP = + u = 2 ms -1 v = ? a = ms -2 s = - 30 m t = ? Worked Example A helicopter travelling upwards at 2 ms -1 releases a box from a height of 30 m. Calculate the landing velocity of the box. v 2 = u 2 + 2as = 2 2 +(2 x x -30) = 592 v = ±24.3 ms -1 v = 24.3 ms -1, down

v 2 = u 2 + 2as Problems to do:33 – 42 on Equations of Motion s = ut + ½ a t 2 v = u + at For all calculations … 1 st Decide a SIGN CONVENTION 2 nd List “u v a s t” 3 rd Select equation(s) and calculate UP = + or - u = v = a = s = t =

v 2 = u 2 + 2as Problems to do:33 – 42 on Equations of Motion s = ut + ½ a t 2 v = u + at m m/s m/s s s 38. a) 750 m/s 2 b) 0.04 s N/kg 40. a) 0.21 m/s 2 b) 1.43 s m 42. a) (i) 21.4 m (ii) 15.6 m/s b) 34.6 m