An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 5 t, - t^{3} + 4 t)$ $v(t) = ( 3t^2 - 5t , -t^3 +4t )$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?

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Answer 1 · 2016-05-30T18:55:55

$a (4) = 33.84 \frac{m}{s^{2}}$ $a(4)=33.84" " m/s^2$

$a_{x} (t) = \frac{d}{d t} (3 t^{2} - 5 t) = 6 t - 5$ $a_x(t)=d/(d t) (3t^2-5t)=6t-5$

$a_{x} (4) = 6 \cdot 4 - 5 = 24 - 5 = 19$ $a_x(4)=6*4-5=24-5=19$

$a_{y} (t) = \frac{d}{d t} (- t^{3} + 4 t) = - 3 t^{2} + 4$ $a_y(t)=d/(d t)(-t^3+4t)=-3t^2+4$

$a_{y} (4) = - 2 \cdot 4^{2} + 4 = - 32 + 4 = - 28$ $a_y(4)=-2*4^2+4=-32+4=-28$

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$a (4) = \sqrt{a_{x} {(4)}^{2} + a_{y} {(4)}^{2}}$ $a(4)=sqrt(a_x(4)^2+a_y(4)^2$

$a (4) = \sqrt{19^{2} + {(- 28)}^{2}}$ $a(4)=sqrt(19^2+(-28)^2)$

$a (4) = \sqrt{361 + 784}$ $a(4)=sqrt(361+784)$

$a (4) = \sqrt{1145}$ $a(4)=sqrt 1145$

$a (4) = 33.84 \frac{m}{s^{2}}$ $a(4)=33.84" " m/s^2$

An object's two dimensional velocity is given by v(t) = ( 3t^2 - 5t , -t^3 +4t )v(t)=(3t2−5t,−t3+4t)v(t) = ( 3t^2 - 5t , -t^3 +4t ). What is the object's rate and direction of acceleration at t=4 t=4t=4 ?