help with rectlinear/curvilinear
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:l thnx for the help
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Not 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 -
thanks, i will discuss it with my friends and get back to you asap
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i seem to have trouble with these 2 practice q's also . they're simple but i keep getting it wrong
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in ref to above, for the second one i just used Pythagoras and like added 90deg to the component to get an angle but it;s wrong . i thought acc would be 90 deg to it
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55 and 52. the negative was just so i could get the direction. ans is in degrees/radians right..?
or are we trying to find out something else
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How did you get the values of 52 and 55?
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oh sorry i was referring to the second picture above, but here it is again http://i105.photobucket.com/albums/m225/nickykeeng/album2/problem.jpg
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I think you post out the question
I can't access pictures from photobucket or any peer sharing image facility at the moment.
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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
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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?
x and y should be referring to the horizontal and vertical components of the velocity.
when t = 0.1667, x = 0.5 - ythus, 5y^2 + 8y = -8x + 7
5y^2 + 8y = -4 + 8y + 7
5y^2 = 3
y = 0.7746 (y is positive)From 5y^2 + 8y = -8x + 7,
10y dy/dt + 8 dy/dt = -8 dx/dt ------- (1)From x + y = 3t
dx/dt + dy/dt = 3
dx/dt = 3 - dy/dt ------- (2)Sub (2) into (1)
10y dy/dt + 8 dy/dt = -24 + 8 dy/dt
dy/dt = -24 / (10y), y = 0.7746
dy/dt = -3.098thus, dx/dt = 3 - dy/dt = 6.098
Thus, v = sqrt {(-3.098)^2 + 6.098^2} = 6.84
Do help me check for careless mistakes... Question 2 I do a while later... I go do something else first.
Regards,
Eagle
Owner of Strategic Tuition -
thanks will do it when i get back. lecture time
ps: the first answer of the 2nd picture is correct (y)
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I think the 2nd question hv to wait for me to be at home then can do
I can't see any graph...
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for that question i would've thought u would just add 90 deg to ur ans
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blur
what do you mean by find the n direction?
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that would be n = normal to the motion . (90 deg heading in), n/t , normal/tangential maybe in other words it's the direction of the instantaneous radius of curvature
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one arrow points up at 55 m/s one arrow points left at -52m /s
I think I remember seeing that the arrow points right at -52 m/s
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sorry meant to say points left at 52
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I still don't understand what is meant by the question in "find the n direction". Because if it is only 2D (and acceleration is still in that 2D plane), it doesn't seem logical to find anything in the normal.
Do you have the answer?
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nah trying to do it hey
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managed to do it
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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..
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friend redid his values and got a small answer of like 1.7... (havent confirmed) why is it soo small!?
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Not sure if I remember correctly...
but I remembered the last term as 8t, not 8x
2y dy/dt = 12x^2 dx/dt + 8
should be 2y dy/dt = 12x^2 dx/dt + 8dx/dt