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A particle moves around a circle (once) such that its

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by BTGmoderatorLU » Sat Jun 30, 2018 4:59 pm
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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

The OA is B.

I'm a little confused by this PS question. Please, can someone give a quick approach to solve it? Thanks in advance!
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RE:

by deloitte247 » Fri Jul 06, 2018 9:43 am
Distance = t [ 6 - t]
Time taken = t

$$Time\ taken\ to\ complete\ \frac{1}{6}\ of\ distance\ $$
$$\frac{1}{6}\ of\ t\ \left[\ 6\ -\ t\right]\ $$
$$\frac{\left[6t\ -\ t^2\right]}{6}$$
$$\frac{\left(\left[6-\ 1\right]\ \left[t^{1-2}\right]\right)}{6}$$
$$\frac{\left(\left[5\ \left[t^{-1}\right]\right]\right)}{6}$$
$$=\ 0.83\ \left[\ t^{-1}\right]$$
t = time in seconds
$$=\ 0.83\ 5^{-1}$$
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