v
acceleration a 

s
t
Rise changein velocity v v f  vi



Run
time taken
t
t
 at  v f  vi

vf = vi + at
x = vf t  ½ a t2
x = vi t + ½ a t2
t
v f  vi
a

ENDY1.4 – Motion: Constant Accelertion
vf 2 = vi 2 + 2ax
x = ½ (vi + vf) t
20  3
 21.3s
0.8
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June 2012
x
v f 2  vi 2
2a

202  32
 244.4m
2  0.8
½
x  vit  0.5  9.8  t 2  50  0  0.5  9.8  t 2  t  3.2s
v f  vi  at  0  9.8  3.2  v f  31ms 1

0  302  2  9.8  x  x  46m




ENDY1.4 – Motion: Constant Accelertion
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June 2012
ENDY1.4 – Motion: Constant Accelertion
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June 2012