An object's two dimensional velocity is given by $v (t) = (\frac{1}{t}, t^{2})$ $v(t) = ( 1/t, t^2)$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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An object's two dimensional velocity is given by $v (t) = (\frac{1}{t}, t^{2})$ $v(t) = ( 1/t, t^2)$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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Answer 1 · 2017-04-26T16:56:52

The acceleration at $t = 2$ $t=2$ is $a (t) = (- \frac{1}{4}, 4)$ $a(t) = (-1/4, 4)$ , whose absolute value is $\frac{\sqrt{257}}{4}$ $sqrt257/4$ and the direction is ${93.576}^{\circ}$ $93.576^@$ w.r.t. $x$ $x$ -axis.

Explanation:

Acceleration is the first derivative of velocity, so if we differentiate the velocity function, we get a function for the acceleration:

$\frac{d v}{d t} = (- \frac{1}{t^{2}}, 2 t)$ $(dv)/(dt) = (-1/t^2, 2t)$

Inserting $t = 2$ $t=2$ gives us the acceleration at $t = 2 s$ $t= 2s$ , namely

$a (t) = (- \frac{1}{4}, 4)$ $a(t) = (-1/4, 4)$ , whose absolute value is $\sqrt{{(- \frac{1}{4})}^{2} + 4^{2}} = \sqrt{\frac{257}{16}} = \frac{\sqrt{257}}{4}$ $sqrt((-1/4)^2+4^2)=sqrt(257/16)=sqrt257/4$

The direction is given by $α = {tan}^{- 1} (\frac{4}{- \frac{1}{4}}) = {tan}^{- 1} (- 16) = {93.576}^{\circ}$ $alpha=tan^(-1)(4/(-1/4))=tan^(-1)(-16)=93.576^@$ w.r.t. $x$ $x$ -axis.

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An object's two dimensional velocity is given by $v (t) = (\frac{1}{t}, t^{2})$ $v(t) = ( 1/t, t^2)$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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