Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

17% (01:28) correct
83% (01:14) wrong based on 23 sessions

HideShow timer Statistics

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

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.

Attachments

Screen shot 2011-05-20 at 2.22.12 AM.png [ 14.54 KiB | Viewed 1974 times ]

_________________

My dad once said to me: Son, nothing succeeds like success.

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.

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

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.
_________________

My dad once said to me: Son, nothing succeeds like success.

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

Attachments

Circle.JPG [ 11.25 KiB | Viewed 1915 times ]

_________________

If you liked my post, please consider a Kudos for me. Thanks!

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...