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

Question

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

Toppr Admin · Accepted Answer

Vavg-Dt-x21D2-0-34-3-06t-x21D2-t-3-060-34 - 9-xA0-secWe have- v-u-atv-x2212-u-at0-18-a-xD7-9-x21D2-a-0-189-0-02-xA0-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

$V_{a v g} = \frac{D}{t} \Rightarrow 0.34 = \frac{3.06}{t}$

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

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

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

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

Vavg=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

$V_{a v g} = \frac{D}{t} \Rightarrow 0.34 = \frac{3.06}{t}$

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

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

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