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04 Jan 2022, 03:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
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80%
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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
(A) 0.60 s
(B) 0.88 s
(C) 1 s
(D) 1.12 s
(E) none of these
(A) 0.60 s
(B) 0.88 s
(C) 1 s
(D) 1.12 s
(E) none of these
06 Mar 2022, 04:56
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Given:
1) Particle moves around a circle.
2) Displacement as function of time (t) = \(t(6-t)\)
Suppose there is a circle having center at O. Particle starts from initial point I.
Now we know that,
when particle moves around a circle, it's displacement will be maximum when it reaches the diametrically opposite point of the initial point.
so, maximum displacement = diameter of circle.
Now see the displacement function. Maximum displacement (say it D) will occur at t = 3. And D = 9.
Circumference of the circle will be the total distance covered by the particle (since it moves around only once). Circumference = \(\pi\) * D
We also know that when particle completes one round, it moves \(360^{\circ}\).
So when particle will move only one sixth of total distance, it will subtend \(60^{\circ}\) at center. Let the final point be F.
Now see the \(\triangle\) IOF. In which OI = OF = Radius. \(\angle\) IOF = \(60^{\circ}\). So we can say that \(\triangle\) IOF is Equilateral Triangle.
So, IF = Radius = 4.5 meters.
IF is the displacement of particle from initial point.
\(t(6-t)=4.5 \)
Solving the above quadratic equation will fetch us, t= 0.88 sec and t= 5.12 sec.
t= 0.88 sec will be the correct answer. Option (B)
at t= 5.12 sec, displacement will be 4.5 meters but Distance covered will not be one-sixth of total distance. In fact, at t= 5.12 sec, distance covered will be five-sixth of total distance.
1) Particle moves around a circle.
2) Displacement as function of time (t) = \(t(6-t)\)
Suppose there is a circle having center at O. Particle starts from initial point I.
Now we know that,
when particle moves around a circle, it's displacement will be maximum when it reaches the diametrically opposite point of the initial point.
so, maximum displacement = diameter of circle.
Now see the displacement function. Maximum displacement (say it D) will occur at t = 3. And D = 9.
Circumference of the circle will be the total distance covered by the particle (since it moves around only once). Circumference = \(\pi\) * D
We also know that when particle completes one round, it moves \(360^{\circ}\).
So when particle will move only one sixth of total distance, it will subtend \(60^{\circ}\) at center. Let the final point be F.
Now see the \(\triangle\) IOF. In which OI = OF = Radius. \(\angle\) IOF = \(60^{\circ}\). So we can say that \(\triangle\) IOF is Equilateral Triangle.
So, IF = Radius = 4.5 meters.
IF is the displacement of particle from initial point.
\(t(6-t)=4.5 \)
Solving the above quadratic equation will fetch us, t= 0.88 sec and t= 5.12 sec.
t= 0.88 sec will be the correct answer. Option (B)
at t= 5.12 sec, displacement will be 4.5 meters but Distance covered will not be one-sixth of total distance. In fact, at t= 5.12 sec, distance covered will be five-sixth of total distance.
kungfury42
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06 Mar 2022, 16:13
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Please don't make this question appear on my actual GMAT, or else I'm going to cry tears of blood.
The answer to this question is option B 0.88s. Here's how:
At any time t, displacement is given by t(6-t)
The displacement vector forms a triangle with the two radii. The central angle of this triangle is given by
Ω = cos^{-1}(2r²-t²(6-t)² ÷ 2r²)
At this point of time, the distance traveled is given by
D = (Ω/360)*2πr = 2πr/6
Therefore, Ω=60
Taking cos on both sides, we get
2r²-t²(6-t)² ÷ 2r² = cos(60) = 1 ÷ 2
Therefore, 2r² - t²(6-t)² = r²
Or t²(6-t)² = r²
Or t(6-t) = r
We have the Displacement function as t(6-t) which can be differentiated to find its maxima. The maxima occurs at 6-2t=0 or at t=3 and is equal to 3*(6-3) = 9 in magnitude.
We know that this will happen at a point which is diametrically opposite to the starting point, and hence 2r = 9 or r = 9/2
Plugging this back into our equation we get,
t(6-t) = r
6t-t² = 9/2
12t-2t² = 9
2t²-12t+9 = 0
Solving this using Quadratic formula we get,
t = 12±√72 ÷ 4
Or t = 12±6√2 ÷ 4
Or t = 3±3/√2
3/√2 is 3/1.41 or ~~2.12
Therefore, t = 0.88 or 5.12
Since we are concerned with the earlier of the two as the concerned point is 1/6th of the circumference away and not 5/6th, our answer here is 0.88s, option B.
Posted from my mobile device
The answer to this question is option B 0.88s. Here's how:
At any time t, displacement is given by t(6-t)
The displacement vector forms a triangle with the two radii. The central angle of this triangle is given by
Ω = cos^{-1}(2r²-t²(6-t)² ÷ 2r²)
At this point of time, the distance traveled is given by
D = (Ω/360)*2πr = 2πr/6
Therefore, Ω=60
Taking cos on both sides, we get
2r²-t²(6-t)² ÷ 2r² = cos(60) = 1 ÷ 2
Therefore, 2r² - t²(6-t)² = r²
Or t²(6-t)² = r²
Or t(6-t) = r
We have the Displacement function as t(6-t) which can be differentiated to find its maxima. The maxima occurs at 6-2t=0 or at t=3 and is equal to 3*(6-3) = 9 in magnitude.
We know that this will happen at a point which is diametrically opposite to the starting point, and hence 2r = 9 or r = 9/2
Plugging this back into our equation we get,
t(6-t) = r
6t-t² = 9/2
12t-2t² = 9
2t²-12t+9 = 0
Solving this using Quadratic formula we get,
t = 12±√72 ÷ 4
Or t = 12±6√2 ÷ 4
Or t = 3±3/√2
3/√2 is 3/1.41 or ~~2.12
Therefore, t = 0.88 or 5.12
Since we are concerned with the earlier of the two as the concerned point is 1/6th of the circumference away and not 5/6th, our answer here is 0.88s, option B.
Posted from my mobile device
( although it has speed distance time concept in this, but still whyyy ??)
