NEET · Physics · STD 11 - 3.2 motion in plane
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 \(0\)
Answer & Solution
Correct Answer
(A) \(-4\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\,\,\,\,\,\,\,x = 5t - 2{t^2},y = 10t\\
\frac{{dx}}{{dt}} = 5 - 4t,\frac{{dy}}{{dt}} = 10\,\,\,\,\,\,\therefore {v_x} = 5 - 4t,{v_y} = 10\\
\frac{{d{v_x}}}{{dt}} = - 4,\frac{{d{v_y}}}{{dt}} = 0\,\,\,\,\,\,\,\,\,\,\therefore {a_x} = - 4,{a_y} = 0\\
Acceleration,\,\vec a = {a_x}\hat i + {a_y}\hat j = 4\hat i\\
\therefore \,The\,acceleration\,of\,the\,particle\,at\,t = 2\,s\\
is\, - 4\,m\,{s^{ - 2}}
\end{array}\)
\,\,\,\,\,\,\,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}\)
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