The average velocity of a body moving with uniform acceleration after travelling a distance of $3.06 m$ is $0.34 m / s$ . The change in velocity of the body is $0.18 m / s$ . During this time, its acceleration is

0.01 {m / s}^{2}

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0.02 {m / s}^{2}

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0.03 {m / s}^{2}

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0.04 {m / s}^{2}

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Solution

The correct option is B $0.02 {m / s}^{2}$
$V_{a v g} = \frac{D}{t} \Rightarrow 0.34 = \frac{3.06}{t}$

$\Rightarrow t = \frac{3.06}{0.34}$ = $9 s e c$

We have, $v = u + a t$
$v - u = a t$
$0.18 = a \times 9$

$\Rightarrow a = \frac{0.18}{9} = 0.02 m / s^{2}$

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The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 m/s. The change in velocity of the body is 0.18 m/s. During this time, its acceleration is

The correct option is B 0.02 m/s2Vavg=Dt⇒0.34=3.06t⇒t=3.060.34 = 9 secWe have, v=u+atv−u=at0.18=a×9⇒a=0.189=0.02 m/s2

The average velocity of a body moving with uniform acceleration after travelling a distance of $3.06 m$ is $0.34 m / s$ . The change in velocity of the body is $0.18 m / s$ . During this time, its acceleration is

The correct option is B $0.02 {m / s}^{2}$
$V_{a v g} = \frac{D}{t} \Rightarrow 0.34 = \frac{3.06}{t}$

$\Rightarrow t = \frac{3.06}{0.34}$ = $9 s e c$

We have, $v = u + a t$
$v - u = a t$
$0.18 = a \times 9$

$\Rightarrow a = \frac{0.18}{9} = 0.02 m / s^{2}$