Question

A body accelerates uniformly from rest at 2m/s².calculate the magnitude of its velocity after travelling 9m.

Answer 1

Answer:

6 m / sec

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Acceleration ( a ) = 2 m / sec² .

Distance ( s ) = 9 m .

We have to find final velocity ( v ) :

From third equation motion we have :

v² = u² + 2 a s

Putting values here we get :

v² = 0² + 2 × 2 × 9

v² = 36

v = √ 36

v = 6

Thus the  magnitude of its velocity is 6 m / sec .

Answer 2

Given:

Acceleration : ${2} m/{s}^2$

Initial velocity : 0

Distance covered : 9m

To Find: Final velocity.

From the third equation of motion

${v}^2 - {u}^2 = 2as$

=> ${v}^2 = 2as + {u}^2$

Now,

Put the values in the equation.

${v}^2 = 2×2×9 + {0}^2$

${v}2 = 36$

v = $\sqrt {36}$

$\therefore{v = 6}$

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${\boxed{\sf\: {V = 6}\: m/s}}}$

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BE BRAINLY!