An object's two dimensional velocity is given by $v (t) = (\sqrt{t^{2} - 1} - 2 t, t^{2} - 5)$ $v(t) = ( sqrt(t^2-1)-2t , t^2-5)$ . What is the object's rate and direction of acceleration at $t = 6$ $t=6$ ?

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An object's two dimensional velocity is given by $v (t) = (\sqrt{t^{2} - 1} - 2 t, t^{2} - 5)$ $v(t) = ( sqrt(t^2-1)-2t , t^2-5)$ . What is the object's rate and direction of acceleration at $t = 6$ $t=6$ ?

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Answer 1 · 2016-06-30T09:24:20

$a (6) = 11.96$ $a(6)=11.96$

$θ = - {85.33}^{o}$ $theta=-85.33 ^o$

Explanation:

$Horizontal component of acceleration :$ $"Horizontal component of acceleration :"$
$a_{x} (t) = \frac{d}{d t} (\sqrt{t^{2} - 1} - 2 t)$ $a_x(t)=d/(d t)(sqrt(t^2-1)-2t)$

$a_{x} (t) = \frac{2 t}{2 \cdot \sqrt{t^{2} - 1}} - 2$ $a_x(t)=(2t)/(2*sqrt(t^2-1))-2$

$a_{x} (6) = \frac{2 \cdot 6}{2 \cdot \sqrt{6^{2} - 1}} - 2$ $a_x(6)=(2*6)/(2*sqrt(6^2-1))-2$

$a_{x} (6) = \frac{6}{\sqrt{35}} - 2$ $a_x(6)=6/sqrt35-2$

$a_{x} (6) = \frac{6 - 2 \cdot \sqrt{35}}{\sqrt{35}} = \frac{- 5.83}{5.92} = - 0.98$ $a_x(6)=(6-2*sqrt35)/sqrt35=(-5.83)/5.92=-0.98$

$Vertical component of acceleration :$ $"Vertical component of acceleration :"$

$a_{y} (t) = \frac{d}{d t} (t^{2} - 5)$ $a_y(t)=d/(d t) (t^2-5)$

$a_{y} (t) = 2 t$ $a_y(t)=2t$

$a_{y} (6) = 2.6 = 12$ $a_y(6)=2.6=12$

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$a (6) = \sqrt{a_{x} {(6)}^{2} + a_{y} {(6)}^{2}}$ $a(6)=sqrt(a_x(6)^2+a_y(6)^2)$

$a (6) \sqrt{{(- 0.98)}^{2} + 12^{2}}$ $a(6)sqrt((-0.98)^2+12^2)$

$a (6) = \sqrt{0.96 + 144}$ $a(6)=sqrt(0.96+144)$

$a (6) = 11.96$ $a(6)=11.96$

$tan θ = \frac{12}{- 0.98}$ $tan theta=12/(-0.98)$

$tan θ = - 12.24$ $Tan theta=-12.24$

$θ = - {85.33}^{o}$ $theta=-85.33 ^o$

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An object's two dimensional velocity is given by $v (t) = (\sqrt{t^{2} - 1} - 2 t, t^{2} - 5)$ $v(t) = ( sqrt(t^2-1)-2t , t^2-5)$ . What is the object's rate and direction of acceleration at $t = 6$ $t=6$ ?

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An object's two dimensional velocity is given by v(t) = ( sqrt(t^2-1)-2t , t^2-5)v(t)=(√t2−1−2t,t2−5)v(t) = ( sqrt(t^2-1)-2t , t^2-5). What is the object's rate and direction of acceleration at t=6 t=6t=6 ?

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An object's two dimensional velocity is given by $v (t) = (\sqrt{t^{2} - 1} - 2 t, t^{2} - 5)$ $v(t) = ( sqrt(t^2-1)-2t , t^2-5)$ . What is the object's rate and direction of acceleration at $t = 6$ $t=6$ ?