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Motion Under Variable Acceleration

A particle mo...

Question

A particle moving on a curve has the position given by $x = f^{'} (t) sin t + f^{''} (t) cos t, y = f^{'} (t) cos t - f^{''} (t) sin t$ at time $t$ where $f$ is a thrice-differentiable function.Then the velocity of the particle at time $t$ is

f^{'''} (t)

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f^{'} (t) + f^{'''} (t)

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f^{'} (t) + f^{''} (t)

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f^{'} (t) - f^{'''} (t)

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Solution

The correct option is B $f^{'} (t) + f^{'''} (t)$
$V_{x}$
$= \frac{d x}{d t}$
$= f^{'} (t) c o s t + f " (t) s i n t - f " (t) s i n t + f "^{'} (t) c o s t$
$= c o s t (f^{'} (t) + f^{'''} (t))$ ...(i)
$V_{y}$
$= \frac{d y}{d t}$
$= - f^{'} (t) s i n t + f " (t) c o s t - f " (t) c o s t - f "^{'} (t) s i n t$
$= - s i n t (f^{'} (t) + f^{'''} (t))$ ....(ii)
Hence
Net velocity
$V_{n e t}$
$= \sqrt{V_{x}^{2} + V_{y}^{2}}$
$= (f^{'} (t) + f^{'''} (t)) \sqrt{s i n^{2} t + c o s^{2} t}$
$= f^{'} (t) + f^{'''} (t)$ .

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