Question

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Answer 1

Answer:

Explanation:

Given:

1. Initial velocity 'u' = 3 m/s due East
2. Final velocity 'v' = 3 m/s due North
3. Time 't' = 6s

Solution:

The angle between Initial velocity and Final velocity (x°) is 90°.

Let the acceleration be 'a' m/s^2

$a = (v - u) \div t$

a = ( 3 m/s north - 3 m/s east )/6

a = ( 3 m/s north + 3 m/s west)/6

a = {√[(3^2)+(3^2)]}/6

a = 3√2/6

a = √2/2

a = 1/√2 m/s^2 due North West (Option (2) )

Hope it helps you

Answer 2

GIVEN :-

• Initial velocity(u) = 3 m/s     (In East direction)

• Final velocity(v) = 3 m/s      (In North direction)

• Time(t) = 6 s

TO FIND :-

The acceleration of the body.

SOLUTION :-

We know,

$\boxed{\dag\ \sf{a=\frac{v-u}{t} }}$

As u and v are perpendicular velocities, then the resultant velocity will be the hypotenuse formed.

Putting the values :-

$\sf{a=\dfrac{\sqrt{3^2+3^2} }{6} }\\\\\sf{\longmapsto a=\dfrac{\sqrt{9+9} }{6} }\\\\\sf{\longmapsto a=\dfrac{\sqrt{18} }{6} }\\\\\sf{\longmapsto a=\dfrac{3\sqrt{2} }{6} }\\\\\sf{\longmapsto a=\dfrac{\sqrt{2} }{2} }\\\\\boxed{\sf{\longmapsto a=\dfrac{1}{\sqrt{2} }\ m/s^2 }\\}$

As the hypotenuse faces north-east, so the answer is :-

3) 1/√2 m/s² towards north-east.