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3-2.Motion in Plane
medium
A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of '$P$' is such that it sweeps out a length $s = t^3+5$, where s is in metres and $t$ is in seconds. The radius of the path is $20\ m$. The acceleration of '$P$' when $t = 2\ s$ is nearly .......... $m/s^2$
A$14$
B$13$
C$12$
D$7.2$
(AIEEE-2010)
Solution
$\begin{array}{l}
s = {t^3} + 5 \Rightarrow velocity,\,v = \frac{{ds}}{{dt}} = 3{t^2}\\
{\rm{Tengential}}\,acceleration\,{a_t} = \frac{{dv}}{{dt}} = 6t\\
Radial\,acceleration\,{a_c} = \frac{{{v^2}}}{R} = \frac{{9{t^4}}}{R}\\
At\,t = 2s,\,{a_t} = 6 \times 2 = 12\,m/{s^2}
\end{array}$
$\begin{array}{l}
{a_c} = \frac{{9 \times 16}}{{20}} = 7.2\,m/{s^2}\\
\therefore {\rm{Resultanl}}\,accelertaion\\
= \sqrt {a_t^2 + a_c^2} = \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}} = \sqrt {144 + 51.84} \\
= \sqrt {195.84} = 14m/{s^2}
\end{array}$
s = {t^3} + 5 \Rightarrow velocity,\,v = \frac{{ds}}{{dt}} = 3{t^2}\\
{\rm{Tengential}}\,acceleration\,{a_t} = \frac{{dv}}{{dt}} = 6t\\
Radial\,acceleration\,{a_c} = \frac{{{v^2}}}{R} = \frac{{9{t^4}}}{R}\\
At\,t = 2s,\,{a_t} = 6 \times 2 = 12\,m/{s^2}
\end{array}$
$\begin{array}{l}
{a_c} = \frac{{9 \times 16}}{{20}} = 7.2\,m/{s^2}\\
\therefore {\rm{Resultanl}}\,accelertaion\\
= \sqrt {a_t^2 + a_c^2} = \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}} = \sqrt {144 + 51.84} \\
= \sqrt {195.84} = 14m/{s^2}
\end{array}$
Standard 11
Physics
