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,

• Correct answer
  A

  $4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$
• B

  $1.2 \mathrm{t}^{2}=4 \mathrm{t}$
• C

  $4 t^{2}=1.2 t$
• D

  $6+4 \mathrm{t}=0.2 \mathrm{t}^{2}$

Solution

Let us draw the figure for given situation,
$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$ $\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$

Hence, the correct answer is option: 'A'.