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

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Answer 1 · 2017-05-06T09:46:33

The rate of acceleration is $=15.01ms^-2$ in the direction of $=177.5º$

The acceleration is the derivative of the velocity.

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

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

Therefore,

$a(4)=(15,cos4)=(15,-0.65)$

The rate of acceleration is

$||a(4)||=sqrt(15^2+(0.65)^2)$

$=sqrt225.4$

$=15.01ms^-2$

The direction is

$theta=arctan(-0.65/15)$

$theta$ lies in the 2nd quadrant

$theta=180-2.5=177.5º$

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