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7. MOTION
medium
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.
Option A
Option B
Option C
Option D
Solution
$u=36 km h ^{-1}=10 m s ^{-1} ; v=0 ; a=? ; S =?$
$(i)$ Applying $\quad v=u+a t$
$0=10+a \times 2$
Therefore, $a=-\frac{10}{2}=-5 m s ^{-2}$
Retardation $=-($ acceleration $)$
$=-\left(-5 m s ^{-2}\right)$
$=+5 m s ^{-2}$
$(ii)$ Applying $v^{2}-u^{2}=2 a S$
$0-(10)^{2}=2 \times(-5) \times S$
$S=\frac{-100}{-10}=10 m$
Standard 9
Science