An object's two dimensional velocity is given by $v (t) = (\sqrt{t - 2} - t, t^{2})$ $v(t) = ( sqrt(t-2)-t , t^2)$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?

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An object's two dimensional velocity is given by $v (t) = (\sqrt{t - 2} - t, t^{2})$ $v(t) = ( sqrt(t-2)-t , t^2)$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?

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Answer 1 · 2016-04-27T09:44:42

$a {(7)}_{x} = - 0, 76$ $a(7)_x=-0,76$
$a {(7)}_{y} = 2 \cdot 7 = 14$ $a(7)_y=2*7=14$
$a (7) = 14, 02$ $a(7)=14,02$
$α = 86, 89^{o}$ $alpha=86,89^o$

Explanation:

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$a {(t)}_{x} = \frac{d}{d t} (\sqrt{t - 2} - t) = \frac{1}{2 \cdot \sqrt{t - 2}} - 1$ $a(t)_x=d/(d t) (sqrt(t-2)-t)=1/(2*sqrt(t-2))-1$

$a {(7)}_{x} = \frac{1}{2 \cdot \sqrt{7 - 2}} - 1$ $a(7)_x=1/(2*sqrt(7-2))-1$

$a {(7)}_{x} = \frac{1}{2 \cdot \sqrt{5}} - 1$ $a(7)_x=1/(2*sqrt5)-1$

$a {(7)}_{x} = \frac{1 - 2 \cdot \sqrt{5}}{2 \cdot \sqrt{5}}$ $a(7)_x=(1-2*sqrt5)/(2*sqrt5)$

$a {(7)}_{x} = \frac{(1 - 2 \cdot \sqrt{5}) \cdot (2 \cdot \sqrt{5})}{(2 \cdot \sqrt{5}) \cdot (2 \cdot \sqrt{5})}$ $a(7)_x=((1-2*sqrt5)*(2*sqrt5))/((2*sqrt5)*(2*sqrt5))$

$a {(7)}_{x} = \frac{2 \cdot \sqrt{5} - 20}{20}$ $a(7)_x=(2*sqrt5-20)/20$

$a {(7)}_{x} = \frac{\sqrt{5} - 10}{10} = \frac{2, 37 - 10}{10} = - 0, 76$ $a(7)_x=(sqrt5-10)/10=(2,37-10)/10=-0,76$

$a {(t)}_{y} = \frac{d}{d t} (t^{2}) = 2 \cdot t$ $a(t)_y=d/(d t) (t^2)=2*t$

$a {(7)}_{y} = 2 \cdot 7 = 14$ $a(7)_y=2*7=14$

$a (7) = \sqrt{{(a {(x)}_{7})}_{7}^{2} + {(a {(7)}_{y})}_{y}^{2}}$ $a(7)=sqrt((a(x)_7)^2+(a(7)_y)^2)$

$a (7) = \sqrt{({(- 0, 76)}^{2} + 14^{2})}$ $a(7)=sqrt(((-0,76)^2+14^2)$

$a (7) = \sqrt{0, 58 + 196}$ $a(7)=sqrt(0,58+196)$

$a (7) = 14, 02$ $a(7)=14,02$

$tan α = \frac{a {(7)}_{y}}{a {(7)}_{x}}$ $tan alpha=(a(7)_y)/(a(7)_x)$

$tan α = \frac{14}{- 0, 76}$ $tan alpha=14/(-0,76)$

$tan α = - 18, 42$ $tan alpha=-18,42$

$α = 86, 89^{o}$ $alpha=86,89^o$

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An object's two dimensional velocity is given by $v (t) = (\sqrt{t - 2} - t, t^{2})$ $v(t) = ( sqrt(t-2)-t , t^2)$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?

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An object's two dimensional velocity is given by v(t) = ( sqrt(t-2)-t , t^2)v(t)=(√t−2−t,t2)v(t) = ( sqrt(t-2)-t , t^2). What is the object's rate and direction of acceleration at t=7 t=7t=7 ?

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An object's two dimensional velocity is given by $v (t) = (\sqrt{t - 2} - t, t^{2})$ $v(t) = ( sqrt(t-2)-t , t^2)$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?