A man runs at a speed of 4 m / s to overtake a standing bus. When he is 6 m behind the door at t=0, the bus moves forward and continuous with a constant acceleration of 1.2 m / ^ 2 . The man reaches the door in time t .

Correct answer: A. 4 t=6+0.6 ^ 2 Let us draw the figure for given situation, 4 t=6+(1)/(2) × 1.2 × ^ 2 4 t=6+0.6 ^ 2

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Physics
Chapter: Motion in One Dimension

BITSAT 2013 · Physics · Motion in One Dimension

A man runs at a speed of 4 m / s to overtake a standing bus. When...

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} / \mathrms^{2}

. The man reaches the door in time

\mathrm{t}

. Then,

A man runs at a speed of 4 m / s to overtake a standing bus. When he is 6 m behind the door at t=0, the bus moves

Options

A $4 \mathrm{t}=6+0.6 \mathrmt^{2}$
B $1.2 \mathrmt^{2}=4 \mathrm{t}$
C $4 t^{2}=1.2 t$
D $6+4 \mathrm{t}=0.2 \mathrmt^{2}$

Correct answer

A. $4 \mathrm{t}=6+0.6 \mathrmt^{2}$

Step-by-step solution

Let us draw the figure for given situation, A man runs at a speed of 4 m / s to overtake a standing bus. When he is 6 m behind the door at t=0, the bus moves

\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrmt^{2}\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrmt^{2}

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