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Running around Circles [#permalink]
 19 May 2011, 11:57
Question Stats:
16% (01:56) correct
83% (01:17) wrong based on 0 sessions
A particle moves around a circle (once) such that its displacement from the initial point in given time t is t(6-t) meters where t is the time in seconds after the start. The time in which it completes one-sixth of the distance is (1) 0.60 s (2) 0.88 s (3) 1 s (4) 1.12 s (5) none of these ------------------------------------------------------ IIM CAT Level Question ------------------------------------------------------
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Last edited by hussi9 on 20 May 2011, 05:21, edited 2 times in total.
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D= t (6-t)
When d = 1/6D, t = ?
I don't think we can solve for t here since we have no idea about the distance.
Ans (5).
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f(x)=6t-t^2 differentiate f'(x)=6-2t Differentiate again f''(x)=-2. Hence maxima will occur at f'(x)=0 0=6-2t t=3 Hence at t=3 f(x)=>f(3)=9 Now f(x) represents maximum displacement. Maximum displacement in a circle is diameter of circle. hence diameter of circle D=9 radius R=4.5 Now full distance will be circumference of circle=2piR= 2pi 4.5=9pi 1/6 of that distance will be 1.5pi Now look at the figure, 1.5pi is the arc of the circle. Central angle will be 60 Hence cord will be 4.5 i.e the displacement is 4.5 when the distance is 1/6th. Now, f(x)=6t-t^2 t^2-6t+4.5=0 Solve for t t=0.88 or t=5.12. hence OA B. Main concept here is to understand difference between displacement and distance. Please let me know if any step is not clear. I have omitted obvious calculations.
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Screen shot 2011-05-20 at 2.22.12 AM.png [ 14.54 KiB | Viewed 926 times ]
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I think this assumption is not correct. What if the question meant just 1/6th of the radius ? The question does not make it obvious. jamifahad wrote: f(x)=6t-t^2
differentiate
f'(x)=6-2t
Differentiate again
f''(x)=-2. Hence maxima will occur at f'(x)=0
0=6-2t
t=3
Hence at t=3 f(x)=>f(3)=9
Now f(x) represents maximum displacement. Maximum displacement in a circle is diameter of circle.
hence diameter of circle D=9 radius R=4.5
Now full distance will be circumference of circle=2piR= 2pi 4.5=9pi
1/6 of that distance will be 1.5pi
Now look at the figure, 1.5pi is the arc of the circle.
Central angle will be 60
Hence cord will be 4.5 i.e the displacement is 4.5 when the distance is 1/6th.
Now, f(x)=6t-t^2
t^2-6t+4.5=0
Solve for t
t=0.88 or t=5.12.
hence OA B.
Main concept here is to understand difference between displacement and distance. Please let me know if any step is not clear. I have omitted obvious calculations.
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gmat1220 wrote: I think this assumption is not correct. What if the question meant just 1/6th of the radius ? The question does not make it obvious. jamifahad wrote: f(x)=6t-t^2
differentiate
f'(x)=6-2t
Differentiate again
f''(x)=-2. Hence maxima will occur at f'(x)=0
0=6-2t
t=3
Hence at t=3 f(x)=>f(3)=9
Now f(x) represents maximum displacement. Maximum displacement in a circle is diameter of circle.
hence diameter of circle D=9 radius R=4.5
Now full distance will be circumference of circle=2piR= 2pi 4.5=9pi
1/6 of that distance will be 1.5pi
Now look at the figure, 1.5pi is the arc of the circle.
Central angle will be 60
Hence cord will be 4.5 i.e the displacement is 4.5 when the distance is 1/6th.
Now, f(x)=6t-t^2
t^2-6t+4.5=0
Solve for t
t=0.88 or t=5.12.
hence OA B.
Main concept here is to understand difference between displacement and distance. Please let me know if any step is not clear. I have omitted obvious calculations. Well, i think there are no assumptions....... this is a great explanation .... thanks and kudos to jamifahad
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gmat1220 wrote: I think this assumption is not correct. What if the question meant just 1/6th of the radius ? The question does not make it obvious.
This is not assumption. Question clearly states that a particle moves around a circle once. hence the distance covered by particle will be circumference of the circle. 1/6 of circumference will be 1.5pi. Also, to solve this under two minutes you do not need any of those calculations. f(x)=6t-t^2 f(0)=0 f(1)=5 f(2)=8 f(3)=9 f(4)=8 f(5)=5 f(6)=0 Now displacement of 5 is at 1 sec. Displacement of 4.5 will be at JUST less than 1 sec.
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The displacement from the starting position can be max when the particle (P) is at diametrically opposite position of S (initial position) => when t(6-t) is max then the value of t(6-t) = 2R where R is the radius of the circle => t = 3 and R = 9/2. Now, when the particles covers 1/6th of the distance => the angle subtended by SP at the center of the circle is 360/6 = 60 degrees SOP Is equilateral triangle => length of SP is R = 9/2 = t(6-t) => t = 0.88
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Re: Running around Circles [#permalink]
20 May 2011, 03:42
good concept,wonder whats the source of this question ?
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Re: Running around Circles [#permalink]
20 May 2011, 03:44
This is a IIM CAT Practice Question
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Re: Running around Circles [#permalink]
20 May 2011, 05:19
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Re: Running around Circles [#permalink]
20 May 2011, 05:25
Updated Question with below TAG: ------------------------------------------------------ IIM CAT Level Question ------------------------------------------------------ Hence forth I will mark them accordingly gurpreetsingh wrote: amit2k9 wrote: good concept,wonder whats the source of this question ? not a gmat question. do not worry about this. _________________
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Re: Running around Circles
[#permalink]
20 May 2011, 05:25
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