Relative Velocity
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a) 25s; 50m
b) 53.1degree, 31.3s
How to do b), especially the time taken part. -
atan(BC/AB) = atan(40m/30m) = 53.1º
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from there,
upwards speed = 1.2 sin 53.1 degrees
time taken = 30 / ( 1.2 sin 53.1 ) = 31.3 s -
how to get part a?
sorry monday exam le, i still so blur -
don't get this part lehs...Originally posted by eagle:from there,
upwards speed = 1.2 sin 53.1 degrees
time taken = 30 / ( 1.2 sin 53.1 ) = 31.3 s
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it is resolving 1.2m/s into vertical and horizontal component.Originally posted by Y_Shun:don't get this part lehs...
The horizontal component no need to bother because rather troublesome to calculate. And if we purely take the vertical distance, we only need to consider the vertical component.
This is the beauty of vectors:
vertical component of velocity only travels in the vertical direction
likewise for horizontal -
I lazy to doOriginally posted by mrcoconut:how to get part a?
sorry monday exam le, i still so blur
Y_Shun, you post your answer for him? -
actually ah, why sin ah?Originally posted by eagle:it is resolving 1.2m/s into vertical and horizontal component.
The horizontal component no need to bother because rather troublesome to calculate. And if we purely take the vertical distance, we only need to consider the vertical component.
i noe it leads to the answer but why v-component is defined by sin not cos when v-component is on the adjacent side? -
actually i know how to do le!Originally posted by eagle:I lazy to do
Y_Shun, you post your answer for him?
just worked it out thanks!
but this is not O level standard right?
like quite easy leh -
i assume the 53.1 degrees is with respect to the horizontal line, and clockwiseOriginally posted by ^tamago^:actually ah, why sin ah?
i noe it leads to the answer but why v-component is defined by sin not cos when v-component is on the adjacent side?
didn't really bother to check what you do... lazy
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w8.....i abit blur....
Edit: nvm i noe which angle liao.. -
omgs i dun understand leh.
mrcoconut:
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yeah same here, but v-component is adjacent to this 53.1º angle.Originally posted by eagle:i assume the 53.1 degrees is with respect to the horizontal line, and clockwise
can't be answer key wrong? -
BAC = 53.1?
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That's wad i got too...........Originally posted by Y_Shun:BAC = 53.1? -
this is O level standard?
why so easy ah?
the diagram is already given or we must draw ourselves? -
how did u get 31.3 secs?Originally posted by mrcoconut:this is O level standard?
why so easy ah?
the diagram is already given or we must draw ourselves? -
i did everything. but i never find the last time nia.Originally posted by lpx88:how did u get 31.3 secs?
but big improvement leh!
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its the last part tht is hard lols. -
The speed of 2.33m/s..shoudn it be the same for both guys?
Assuming speed of river is constant
And they have the same speed in still water? -
.........................................................................Originally posted by mrcoconut:i did everything. but i never find the last time nia.
but big improvement leh!
Chey -.- I tot u inhuman sia..last part also no kick -
now then I check your answerOriginally posted by ^tamago^:yeah same here, but v-component is adjacent to this 53.1º angle.
can't be answer key wrong?
you got the correct answer but the method is wrong
swimming at angle teta does not mean swimming directly to c
furthermore, teta should be wrt to the horizontal
means
Rindy aims at a location behind B and starts swimming towards there. That's why he will end up at C instead of D -
btw, which year is this question from? I assume it is an O level question
My method uses the R-formula, but I think it is already out of syllabus
I try and see got any other method -
not sure if its 'adapted from o lvl', its from a school exam prelim paper...>_>
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Which sch?Originally posted by Y_Shun:not sure if its 'adapted from o lvl', its from a school exam prelim paper...>_>