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A particle starts moving along a line from zero initial velocity and comes to rest after moving distance d. During its motion it had a constant acceleration f over $2 / 3$ of the distance, and covered the rest of the distance with constant retardation. The time taken to cover the distance is
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Verified Answer
The correct answer is:
$\sqrt{3 \mathrm{~d} / \mathrm{f}}$
$\mathrm{a}_{1} l_{1}=\mathrm{a}_{2} l_{2}$
by equation of motion
$\begin{array}{l}
\text { Retardation }=2 f \\
t=t_{1}+t_{2} \\
=\sqrt{\frac{4 d}{3 f}}+\sqrt{\frac{2 d}{2 f}}=\sqrt{\frac{3 d}{f}}
\end{array}$
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