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

Question

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

Tardigrade Admin · Accepted Answer

Time taken in covering a distance of 3.06 m is textt =( textDistance/ textAverage velocity)=(3.06/0.34)=9s Also, change in velocity v-u=0.18 ms-1 From first law of motion v=u+at ⇒ a=(v-u/t)=(0.18/9)=0.02 ms-2

Q. The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06m 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:

$0.01 m s^{- 2}$

$0.02 m s^{- 2}$

$0.03 m s^{- 2}$

$0.04 m s^{- 2}$

Time taken in covering a distance of 3.06 m is $t = \frac{Distance}{Average velocity} = \frac{3.06}{0.34} = 9 s$ Also, change in velocity $v - u = 0.18 m s^{- 1}$ From first law of motion $v = u + a t$ $\Rightarrow$ $a = \frac{v - u}{t} = \frac{0.18}{9} = 0.02 m s^{- 2}$

Q. The average velocity of a body moving with uniform acceleration after 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