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A body starting from rest and moving with a constant acceleration covers a distance $s_{1}$ in $4 secs$ and covers a distance $s_{2}$ in $6 secs$ . The ratio $\frac{s_{1}}{s_{2}}$ is

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\frac{2}{3}

\frac{4}{9}

\frac{6}{11}

\frac{7}{11}

answer is B.

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Detailed Solution

A body starting from rest and moving with a constant acceleration covers a distance

s_{1}

4 secs

and covers a distance

s_{2}

6 secs

. The ratio

\frac{s_{1}}{s_{2}}

\frac{4}{9}

.
Given
Initial speed

u = 0

The time taken,

t = 4 s

Let

a

be the acceleration of the given body.
Now,

s_{1} = ut + \frac{1}{2} a t^{2}

s_{1} = \frac{1}{2} a {(4)}^{2}

s_{1} = 8 a . . . (1)

And

s_{2} = ut + \frac{1}{2} a t^{2}

s_{2} = \frac{1}{2} a {(6)}^{2}

s_{2} = 18 a . . . (2)

Taking the ratio of equation

(1)

and equation

(2)

, we get:

\frac{s_{1}}{s_{2}} = \frac{8 a}{18 a}

= > \frac{s_{1}}{s_{2}} = \frac{4}{9}

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