Question

An object travels at 3.06 m distance with uniform acceleration in this interval it's velocity increases by 0.18 and it's average velocity over the interval was 0.34 m/s what is the acceleration?

Answer 1

Given that,

Distance travelled, d = 3.06 m

Final velocity, v = 0.18 m/s

Average velocity of the object, V = 0.34 m/s

To find,

The acceleration of the object.

Solution,

Let t is the time in which it covers distance d. Average velocity is given by distance covered divided by time.

`V=(d)/(t)\\\\t=(d)/(V)\\\\t=(3.06)/(0.34)\\\\t=9\ s`

Let a is the acceleration of the object. Using equation of motion to find it :

`a=(v-u)/(t)\\\\a=(0.18-0)/(9)\\\\a=0.02\ m/s^2`

So, the acceleration of the object is 0.02 m/s².