An object's two dimensional 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 $t = 5$ $t=5$ ?

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Answer 1 · 2016-05-11T19:26:40

$a (5) = 72.56$ $a(5)=72.56$
$angle= {69.44}^{o}$ $"angle="69.44^o$

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

$for t= 5; a_{x} (5) = 2.5 - 1 = 10 - 1 = 9$ $"for t="5 " ; "a_x(5)=2.5-1=10-1=9$

$a_{x} (5) = 9$ $a_x(5)=9$

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

$for t=5 ; a_{y} (5) = 3 \cdot 5^{2} - 3 = 75 - 3$ $"for t=5 ;"a_y(5)=3*5^2-3=75-3$

$a_{y} (5) = 72$ $a_y(5)=72$

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$a (5) = \sqrt{72^{2} + 9^{2}} a (5) = \sqrt{5265}$ $a(5)=sqrt(72^2+9^2)" "a(5)=sqrt(5265)$

$a (5) = 72.56$ $a(5)=72.56$

An object's two dimensional 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=5t=5 ?