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

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Answer 1 · 2017-04-29T16:05:54

The rate of acceleration is $= 5.04 m s^{- 2}$ $=5.04ms^-2$ in the direction $= 172.6$ $=172.6$ º

Acceleration is the derivative of the velocity

$a(t)=v'(t)$

$v(t)=(t^2-3t+2,sint)$

$a(t)=v'(t)=(2t-3,cost)$

When $t=4$ ,

$a(4)=(2*4-3,cos4)=(5,-0.65)$

The rate of acceleration is

$||a(4)||=sqrt((5)^2+(-0.65)^2)$

$=sqrt25.42=5.04$

The direction is in the 2nd quadrant

$theta=180-arctan(0.65/5)=172.6º$

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