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the working is a little
.and hard to type out lol,
but the correct answers are1) 9.423 m/s^2
2) 16.593 m/sfor the second one you have to:
differentiate equation – (1)
substitute y= 11 into the equation to get x
substitute the x you found above into (1) – (2)
tan theta = (2)it’s all angle work, similar triangles and stuff in between here.
eventually you take a small angle away from a big angle to get the angle with the velocity (the gradient at y = 11)cos (angle with velocity) = (r x theta dot) / v
theta dot is your rev/min converted to rad/sec
r is found by using trig
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the second question looks simple but it’s not that simple.
take note the thing is not at origin.
we found r using Pythagoras, and theta dot.
r dot cannot be zero.v = r x theta dot does not work because we cant assume r dot to be zero.
we’re given an equation, i was thinking differentiating that to find the gradient of velocity, and thus finding the angles using inverse tan, but what do we do with that?
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i got 25.588 m/s^2.... not sure if it's right or not ._. my d2y/dt2 was negative 24.6126!! ??
tomorrow is the d-day ._.
update: managed to do it... the differentiation chain rule is wrong for the second part, and what we've found is "overall" acceleration. so needed to find the components using trig
Edited by WoAiMeiMei 11 Aug `08, 3:37PM
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Originally posted by eagle:
ot sure if I understood the questions correctly.
First question
At 5s, speed = 3.2*5 = 16m/s
And centripetal acceleration at 16m/s = v^2/r = 14.222222 m/s^2To find magnitude of acceleration when t=5, use pygathoras theroem.
magnitude = sqrt(14.222222^2 + 3.2^2) = 14.58 m/s^22nd question
When x=3, y = sqrt(4*3^3 +8*3) = 11.489
differentiate w.r.t. time
2y dy/dt = 12x^2 dx/dt + 8
When x=3, dx/dt=7, y=11.489 => dy/dt = 33.249differentiate w.r.t. time
2y d2y/dt2 + 2 (dy/dt)^2 = 12x^2 d2x/dt2 + 24x (dx/dt)^2
when x=3, dx/dt=7, d2x/dt2=7, y=11.489, dy/dt = 33.249
d2y/dt2 = 90.217Hence, magnitude of normal component of acceleration = sqrt(90.217^2 + 7^2) = 90.5 m/s^2
Not sure if this is what you want.
Regards,
Eagle
Owner of Strategic Tuition
unfortunately the answer to the last question did not agree. i dun see whats wrong though. still working on it2y dy/dt = 12x^2 dx/dt + 8
should be 2y dy/dt = 12x^2 dx/dt + 8dx/dt
UPDATE: *ignore everything below* friend's answer is wrong
seems that my friend's value was really small... <1, his actual question values were almost half of mine.
His question was:
y^2 = 5x^3 + 5x
x=4
x=3m/s
x=4m/s^2His answer was:
0.7704 m/s^2
it seems your method is used to find the overall acceleration.
apparently you gotta put them all into its components
ie. tangential acceleration and the normal accelerationdun get it..
Edited by WoAiMeiMei 07 Aug `08, 10:06PM
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is is used when you're talking about singular terms eg. He is, She is, it is, the cat is, my mum is, my teacher is are is used when you're talking about plural terms. eg. They are, my friends are, the teachers are, my cats are.
for ur question i reckon all of them would be are since the are always at least 2 subjects
Edited by WoAiMeiMei 06 Aug `08, 10:48AM
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1) Given that
5y^2 + 8y = -8x + 7 and
x + y = 3t
where x and y are in metres and t is in seconds; If y is positive, what is the magnitude of the velocity when t = 0.1667s?
2) the second one:
At some time t, a particle has the velocity components shown. If the acceleration at this time is in the positive x direction, find the n direction with respect to the x axis.
the graph shows a normal positive x/y axis and on it there are two arrows from a reference point. offtopic: what's with the new posting rule or something, it's like i cant write anything unless i click "visual"... one arrow points up at 55 m/s one arrow points left at -52m /s
Edited by WoAiMeiMei 06 Aug `08, 10:41AM
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:O
