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Find the velocity, acceleration, and speed of a particle with the given position function. r(t) = t i t2 j 2 k

Sagot :

The velocity of a particle = i + 2t j

The acceleration of a particle = 2 j

The speed of a particle = [tex]\sqrt{1 + 4t^{2} }[/tex]

Here,

The position function is, [tex]r (t ) = ti + t^{2} j +2k[/tex]

We have to find, the velocity, acceleration, and speed of a particle with the given position function.

What is Velocity of a particle with the given position function?

The instantaneous velocity v(t) of a particle is the derivative of the position with respect to time. That is, v(t)=dx/dt.

Now,

The position function is, [tex]r (t ) = ti + t^{2} j +2k[/tex]

The velocity of a particle = [tex]\frac{d r(t)}{dt}[/tex]

[tex]r (t ) = ti + t^{2} j +2k[/tex]

[tex]\frac{d r(t)}{dt} = i + 2t j[/tex]

The acceleration of a particle = [tex]\frac{d^{2} r(t)}{dt^{2} }[/tex]

[tex]r (t ) = ti + t^{2} j +2k[/tex]

[tex]\frac{d r(t)}{dt} = i + 2t j[/tex]

[tex]\frac{d^{2} r(t)}{dt^{2} }= 2j[/tex]

The speed of a particle = [tex]| \frac{d r(t)}{dt}| = |v(t)|[/tex]

[tex]\frac{d r(t)}{dt} = i + 2t j[/tex]

[tex]|v(t)|=\sqrt{1 + 4t^{2} }[/tex]

Hence, The velocity of a particle = [tex]i + 2t j[/tex]

The acceleration of a particle = [tex]2 j[/tex]

The speed of a particle = [tex]\sqrt{1 + 4t^{2} }[/tex]

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