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3-2.Motion in Plane
medium
The $x$ and $y$ coordinates of the particle at any time are $x = 5t - 2t^2$ and $y = 10t$ respectively, where $x$ and $y$ are in metres and $t$ in seconds. The acceleration of the particle at $t = 2\, s$ is......$m/sec^2$
A
$-4$
B
$-5$
C
$-8$
D
$0$
(NEET-2017)
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,x = 5t – 2{t^2},y = 10t\\
\frac{{dx}}{{dt}} = 5 – 4t,\frac{{dy}}{{dt}} = 10\,\,\,\,\,\,\therefore {v_x} = 5 – 4t,{v_y} = 10\\
\frac{{d{v_x}}}{{dt}} = – 4,\frac{{d{v_y}}}{{dt}} = 0\,\,\,\,\,\,\,\,\,\,\therefore {a_x} = – 4,{a_y} = 0\\
Acceleration,\,\vec a = {a_x}\hat i + {a_y}\hat j = 4\hat i\\
\therefore \,The\,acceleration\,of\,the\,particle\,at\,t = 2\,s\\
is\, – 4\,m\,{s^{ – 2}}
\end{array}$
Standard 11
Physics
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