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A man runs at a speed of $4 \mathrm{~m} / \mathrm{s}$ to overtake a standing bus. When he is $6 \mathrm{~m}$ behind the door at $\mathrm{t}=0,$ the bus moves forward and continuous with a constant acceleration of $1.2 \mathrm{~m} / \mathrm{s}^{2}$. The man reaches the door in time $\mathrm{t}$. Then,
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Verified Answer
The correct answer is:
$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$
Let us draw the figure for given situation,
$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$
$\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$
$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$
$\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$
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