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Ainat [17]
3 years ago
7

A different particle moves in 3 dimensional space. Its velocity is v = {t^2i + 2t^2j + 3 rk} in [m/s]. Find the magnitude of its

acceleration at t = 3 s
Physics
1 answer:
hjlf3 years ago
8 0

Explanation:

It is given that,

Velocity of the particle is given by:

v=(t^2i+2t^2j+3tk)

We need to find the magnitude of its acceleration at t = 3 seconds. We know that, acceleration a is given by :

a=\dfrac{dv}{dt}

a=\dfrac{d(t^2i+2t^2j+3tk)}{dt}            

a=(2t\ i+4t\ j+3\ k)

At t = 3 seconds

a=(2(3)\ i+4(3)\ j+3\ k)

a=6\ i+12\ j+3\ k

So, the magnitude of a is given by :

|a|=\sqrt{6^2+12^2+3^2}

a=13.7\ m/s^2

So, the magnitude of acceleration is 13.7 m/s². Hence, this is the required solution.

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