An object's two dimensional velocity is given by $v (t) = (t - 2, 5 t^{2} - 3 t)$ $v(t) = ( t-2 , 5t^2-3t)$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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Answer 1 · 2016-05-26T15:49:43

$a = 17.03 \frac{m}{s^{2}}$ $a=17.03" "m/s^2$

$tan α = 17$ $tan alpha=17$

$v (t) = (t - 2, 5 t^{2} - 3 t)$ $v(t)=(t-2,5t^2-3t)$

$\frac{d v (t)}{d t} = a (t)$ $(d v(t))/(d t)=a(t)$

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

$a_{y} (t) = \frac{d}{d t} (5 t^{2} - 3 t) = 10 t - 3$ $a_y(t)=d/(d t)(5t^2-3t)=10t-3$

$a_{y} (2) = 10 \cdot 2 - 3$ $a_y(2)=10*2-3$

$a_{y} (2) = 17$ $a_y(2)=17$

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

$a = \sqrt{1^{2} + 17^{2}}$ $a=sqrt(1^2+17^2)$

$a = 17.03 \frac{m}{s^{2}}$ $a=17.03" "m/s^2$

$tan α = 17$ $tan alpha=17$

An object's two dimensional velocity is given by v(t)=(t−2,5t2−3t)v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=2t=2 ?