Asked by Seyifunmi
determine the average acceleration of a slalom skier if she is moving at a velocity of 9.75 m/s[S25E] and then changes her velocity of 8.64 m/s [W35S] in a matter of 0.179sec.
Answers
Answered by
Damon
original south = 9.75 cos 25
final south = 8.64 sin 35
original east = 9.75 sin 25
final east = -8.64 cos 35
south a component = (final south -original south)/ .179
east a component = (final east component - original east component) / .179
resultant mag of a = sqrt (south a^2 + east a^2 )
angle T south of east
tan T = final a south / final a east
final south = 8.64 sin 35
original east = 9.75 sin 25
final east = -8.64 cos 35
south a component = (final south -original south)/ .179
east a component = (final east component - original east component) / .179
resultant mag of a = sqrt (south a^2 + east a^2 )
angle T south of east
tan T = final a south / final a east
Answered by
bobpursley
acceleration=(vfinal-Vinitial)/time
Now, notice the velocity is vectors. I would do it like this
vf=9.75cos25 S + 9.75sin25 E
Vi=8.64sin35 S + 8.64cos35 W
remember W is -E, so when you subtract..
Vf-Vi=(9.75cos25 -8.64sin35)S +(9.75sin25+8.64cos35) E
do that math, divide by time
If you wish, change it to polar form, the problem did not require it.
Now, notice the velocity is vectors. I would do it like this
vf=9.75cos25 S + 9.75sin25 E
Vi=8.64sin35 S + 8.64cos35 W
remember W is -E, so when you subtract..
Vf-Vi=(9.75cos25 -8.64sin35)S +(9.75sin25+8.64cos35) E
do that math, divide by time
If you wish, change it to polar form, the problem did not require it.
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