A cyclist driving at 36 , km , h^{-1} stops h | vedclass

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(A) Given: Initial velocity $u = 36 \, km \, h^{-1} = 36 	imes \frac{5}{18} = 10 \, m \, s^{-1}$.Final velocity $v = 0 \, m \, s^{-1}$ (since the cyc

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$5 \, m \, s^{-2}, 10 \, m$

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(A) Given: Initial velocity $u = 36 \, km \, h^{-1} = 36 	imes \frac{5}{18} = 10 \, m \, s^{-1}$.Final velocity $v = 0 \, m \, s^{-1}$ (since the cycle stops).Time $t = 2 \, s$.$(i)$ To calculate retardation:Using the first equation of motion,$v = u + at$:$0 = 10 + a 	imes 2$$2a = -10$$a = -5 \, m \, s^{-2}$.Retardation is the negative of acceleration,so retardation $= -(-5 \, m \, s^{-2}) = 5 \, m \, s^{-2}$.$(ii)$ To calculate distance covered $(S)$:Using the third equation of motion,$v^2 - u^2 = 2aS$:$0^2 - (10)^2 = 2 	imes (-5) 	imes S$$-100 = -10 	imes S$$S = \frac{-100}{-10} = 10 \, m$.

$A$ cyclist driving at $36 \, km \, h^{-1}$ stops his cycle in $2 \, s$ by the application of brakes. Calculate $(i)$ retardation $(ii)$ distance covered during the application of brakes.

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