An object's velocity is given by $v (t) = (t^{2} - t + 1, t^{3} - 3 t)$ $v(t) = (t^2 -t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t = 6$ $t=6$ ?

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Answer 1 · 2016-05-06T21:35:26

$a (6) = 105.57$ $a(6)=105.57$
$θ = {84.02}^{o}$ $theta=84.02^o$

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$a_{x} (t) = \frac{d}{d t} = (t^{2} - t + 1)$ $a_x(t)=d/(d t)=(t^2-t+1)$

$a_{x} (t) = 2 t - 1 a_{x} (6) = 2 \cdot 6 - 1 = 11$ $a_x(t)=2t-1" "a_x(6)=2*6-1=11$

$a_{y} (t) = \frac{d}{d t} (t^{3} - 3 t)$ $a_y(t)=d/(d t)(t^3-3t)$

$a_{y} (t) = 3 t^{2} - 3 a_{y} (6) = 3 \cdot 6^{2} - 3 = 108 - 3 = 105$ $a_y(t)=3t^2-3" "a_y(6)=3*6^2-3=108-3=105$

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

$a (6) = \sqrt{11^{2} + 105^{2}}$ $a(6)=sqrt(11^2+105^2)$

$a (6) = \sqrt{121 + 11025}$ $a(6)=sqrt(121+11025)$

$a (6) = 105.57$ $a(6)=105.57$

$tan θ = \frac{a_{y} (6)}{a_{x} (6)} = \frac{105}{11}$ $tan theta=(a_y(6))/(a_x(6))=105/11$

$θ = {84.02}^{o}$ $theta=84.02^o$

An object's velocity is given by v(t)=(t2−t+1,t3−3t)v(t) = (t^2 -t +1 , t^3- 3t ). What is the object's rate and direction of acceleration at t=6t=6 ?