The average velocity of a body moving with uniform acceleration after 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 after 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 ms -2 (B) 0.02 m s -2 (C) 0.03 m s -2 (D) 0.04 m s -2

Tardigrade Admin · Accepted Answer

Given. v av =(v+u/2)=0.34 and v-u=0.18 Solving these two equations, we get u=0.25 ms -1, v=0.43 ms -1. Given s=3.06 m Now use v2-u2=2 a s to find a=0.02 ms -2.

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

$0.01 m s^{- 2}$

$0.02 m s^{- 2}$

$0.03 m s^{- 2}$

$0.04 m s^{- 2}$

Given. $v_{a v} = \frac{v + u}{2} = 0.34$ and $v - u = 0.18$
Solving these two equations,
we get $u = 0.25 m s^{- 1}, v = 0.43 m s^{- 1}$ .
Given $s = 3.06 m$
Now use $v^{2} - u^{2} = 2 a s$ to find $a = 0.02 m s^{- 2}$ .

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