The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 m s- 1 . If the change in velocity of the body is 0.18 m s- 1 during this time, its uniform acceleration is

Question

The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 m s- 1 . If the change in velocity of the body is 0.18 m s- 1 during this time, its uniform acceleration is (A) 0.01 m s- 2 (B) 0.02 m s- 2 (C) 0.03 m s- 2 (D) 0.04 m s- 2

Tardigrade Admin · Accepted Answer

Average velocity =(D i s p l a c e m e n t/t i m e) 0.34=(3.06/t i m e) time =(3.06/0.34)sec Acceleration =(c h ange â¡ in â¡ v e l o c i t y/t i m e) Acceleration =(0.18/((3.06)0.34)) =0.02 m s- 2

Q. The average velocity of a body moving with uniform acceleration travelling a distance of $3.06 m$ is $0.34 m s^{- 1}$ . If the change in velocity of the body is $0.18 m s^{- 1}$ during this time, its uniform acceleration is

$0.01 m s^{- 2}$

$0.02 m s^{- 2}$

$0.03 m s^{- 2}$

$0.04 m s^{- 2}$

Average velocity $= \frac{D i s pl a ce m e n t}{t im e}$
$0.34 = \frac{3.06}{t im e}$
$t im e = \frac{3.06}{0.34} sec$
$A cce l er a t i o n = \frac{c han g e in v e l oc i t y}{t im e}$
$A cce l er a t i o n = \frac{0.18}{( \frac{3.06}{0.34} )}$
$= 0.02 m s^{- 2}$

Q. The average velocity of a body moving with uniform acceleration travelling a distance of 3.06m is 0.34ms−1 . If the change in velocity of the body is 0.18ms−1 during this time, its uniform acceleration is

0.01ms−2

0.02ms−2

0.03ms−2

0.04ms−2