A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is

0, 1

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\frac{π + 4}{2}

, 0

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21.4,

\frac{π + 4}{2}

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0,21.4

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Solution

The correct option is B 0,21.4
Since the initial position coincides with the final position. So, the net displacement of the cyclist = zero
Average speed of the cyclist = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n}$

$= \frac{O P + P Q + Q O}{10} k m m i n^{- 1}$

$= \frac{π + 4}{20} \times 60 k m h^{- 1} = 21.4 k m h^{- 1}$

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The correct option is B 0,21.4Since the initial position coincides with the final position. So, the net displacement of the cyclist = zeroAverage speed of the cyclist = Total distance travelledTotal time taken=OP+PQ+QO10km min−1 =π+420×60kmh−1=21.4kmh−1