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A cyclist riding with a speed of 27 kmph. As he approaches a circular turn on the road of radius 80 m, he applies breaks and reduces his speed at the constant rate of 0.50 m/s every second. The net acceleration of cyclist on the circular turn is

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a

0.5 m/s²

b

0.8 m/s²

c

0.86 m/s²

d

1 m/s²

answer is C.

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Detailed Solution

$\begin{array}{l} a_{net} = \sqrt{a_{c}^{2} + a_{t}^{2}} \\ a_{c} = \frac{v^{2}}{r} = \frac{{(27 \times \frac{5}{18})}^{2}}{80} = \frac{225}{4 \times 80} = \frac{45}{64} = 0.70 \end{array}$
a_t = 0.5
$a_{net} = \sqrt{(0.7)^{2} + (0.5)^{2}} = \sqrt{0.74} = 0.86 {ms}^{- 2}$

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