Relative Velocity
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presbyterian secOriginally posted by lpx88:Which sch? -
ok I got a method which you will have learned before
but u all try a bit first hor
Small Hint:
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i thought of that but i have weird vectors. i actually went there one round before concluding to try test atan(0.8). i go try again.Originally posted by eagle:now then I check your answer
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
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Thx eagle.. I got the answer...ive not drawn such a velocity diagram before though, which probably explains y i cant get the ans
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50 / sqrt(4 + (1.2 * 1.2) + (4.8 * cos((53.1 / 180) * pi))) = 17.33sOriginally posted by eagle:ok I got a method which you will have learned before
but u all try a bit first hor
Small Hint:
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Using sine rule, u shud get that big angle red as a right angle.Originally posted by ^tamago^:50 / sqrt(4 + (1.2 * 1.2) + (4.8 * cos((53.1 / 180) * pi))) = 17.33s
With that get angle in green..180 degrees - 90-36.9 = 53.1
Cause right angle triangle, can use Pythagoras theorem to get velocity the guy travelling at... sqrt (2 sqaure - 1.2 sqaure)
Distance is 50m from another Pythagoras theorem as well. Sqrt (40 square + 30 square)
Dist/velocity=31.3 sec(Rounded off) -
I hope everyone knows why it is 36.9 degreesOriginally posted by ^tamago^:50 / sqrt(4 + (1.2 * 1.2) + (4.8 * cos((53.1 / 180) * pi))) = 17.33s
then:
find alpha using sine rule (ans: 90 degrees)
then you can find teta -
this part can use my earlier solutionOriginally posted by lpx88:Cause right angle triangle, can use Pythagoras theorem to get velocity the guy travelling at... sqrt (2 sqaure - 1.2 sqaure)
Distance is 50m from another Pythagoras theorem as well. Sqrt (40 square + 30 square)
Dist/velocity=31.3 sec(Rounded off) -
50 / sqrt(4 + (1.2 * 1.2) - (4.8 cos((53.1 / 180) * pi))) = 31.2623186Originally posted by ^tamago^:50 / sqrt(4 + (1.2 * 1.2) + (4.8 * cos((53.1 / 180) * pi))) = 17.33s
walao waste time. could have solved it in much earlier. bang my own balls. -
thanks all for the help. it's a honest arithmetic mistake.
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i dun get it =x.
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AC is 50m. Find the resultant velocity (in green) from A to C through the diagram and use AC divide by the speed. -
How come AC is 50?
I thought its BD? -
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ABC is a right-angled triangle.Originally posted by Y_Shun:How come AC is 50?
I thought its BD?
AB=30m
BC=40m
So hyp AC=50m
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how you get the 90degree there? i get a 'math error'...
2/sinx = 1.2/sin 36.9 -
er. move a bit earlier. ignore the right-angle and try to find all the angles involved. see eagle's drawing.Originally posted by Y_Shun:how you get the 90degree there? i get a 'math error'...
2/sinx = 1.2/sin 36.9 -
ya i trying to find his red angle, but can't!Originally posted by ^tamago^:er. move a bit earlier. ignore the right-angle and try to find all the angles involved. see eagle's drawing. -
er, it's 90º.Originally posted by Y_Shun:ya i trying to find his red angle, but can't!
actually the main thing is the 53.1º leh.
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but but but...how you get tht 90? >.<Originally posted by ^tamago^:er, it's 90º.
actually the main thing is the 53.1º leh.
i tot u need to use
180 - 36.9 - 90 = 53.1 ? -
Using sine ruleOriginally posted by Y_Shun:but but but...how you get tht 90? >.<
i tot u need to use
180 - 36.9 - 90 = 53.1 ?
But u don use 36.9
Coz its a rounded number
Use yr calculator for atan(30/40) to get 36.8 sth
Use that value for sine rule -
Final qn..can some1 remind me about the decimal points and sig fig for a maths?
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----------- another qtn
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the new qns like this yr june paper a maths rela velo qns leh
if im nt wrong lol
i try bah...
6(i)
Main point here : when they intersect (thats when the missile hits the spacecraft), their coordinates are the same.
so,
let the time taken from the time when the missile is launched to the time when it hits the spacecraft be t
to find the final coordinates of the spacecraft and missile, u have to take their position vectors plus their speed*time taken. (for spacecraft add 0.5 sec more to time taken since missile was only launched 0.5 sec after it)
SPACECRAFT MISSILE
12j + (t+0.5)(40i + 15j) = 46i + t(ki + 30j)
12j + 20i + 7.5j + (40t)i + (15t)j = 46i + (tk)i + (30t)j
(20+40t)i + (19.5+15t)j = (46+tk)i + (30t)j
19.5+15t = 30t
19.5 = 15t
t=1.3
Therefore, time taken for spacecraft to move across screen before impact = 1.3+0.5 = 1.8s [shown]
(ii)
20+40t = 46+tk
Sub t=1.3
20+52 = 46+1.3k
k = 20 -
There's a simliar question
(same diagram), which is not drawn to scale, shows a horizontal rectangular surface. One corner of the surface is taken as the origin O and i and j are unit vectors along the edges of the suface.
A bloody fly, F, starts at a point with position vector (i+2j) and crawls across the surface with a velocity of (3i + 2j) cm s-1, At the instant that bloody fly starts crawling, a more bloody spider, S, at the point with position vector (85i + 5j) cm, sets off across the surface wit ha velocity of (-5i + kj)cm/s, where k = constant, given that the bloody insects knock on each other, find k.
heres my working
After t seconds, the bloody fly travels : 3it + 2jt cm
the bloody spider : kjt - 5it
since the spider catches the fly, the distance they travelled after t s is the same
3it + 2jt + i + 12j = 85i + 5j - 5it + kjt
Simplifying, i get : i(8t - 84) = j(-7+kt - 2t)
By whatever rule is that, i forgot, 8t - 84 = 0
t = 10.5
-7 + 10.5k - 21 = 0
k = 8/3
This same method can be applied for the question above, I'll do it if i got time, but i wonder, is it RV or velocity?