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

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Answer 1 · 2017-06-10T11:32:30

The rate of acceleration is $= 11.04 m s^{- 2}$ $=11.04ms^-2$ in the direction $=5.14º$ clockwise from the x-axis

The acceleration is the derivative of the velocity

$v(t)=(2t^2-t+2,sint)$

$a(t)=v'(t)=(4t-1,cost)$

So,

when $t=3$

$a(3)=(11,cos3)=(11,-0.99)$

The rate of acceleration is

$=||a(3)||=sqrt(11^2+(-0.99)^2)=sqrt121.98=11.04$

The direction is $arctan(-0.99/11)=-5.14º$

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