Question

The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 m
If the change in velocity of the body is 0.18ms during this time, its uniform acceleration is
(a) 0.01 ms
(b) 0.02 ms?
(c) 0,03 ms?
(d)0,04 ms​

Answer 1

Answer:

Option-b-0.02m/s^2

Explanation:

Given that, v=0.34m/s

d=3.06m

Change in v=0.18m/s

$v = 0.34 = \frac{distance}{time} = \frac{3.06}{t} = \frac{3.06}{ \frac{change \: in \: v}{a} } = \frac{3.06 \times a}{0.18} \\ a = \frac{0.34 \times 0.18}{3.06} = 0.02 \frac{m}{ {s}^{2} }$