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An object's velocity is given by $v(t) = (t^2 -t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t=2$ ?

1 Answer

Chitram Banerjee

May 3, 2016

$v_x(t) = t^2-t+1$

$a_x(t) = dotv_x(t) = 2t-1$

$:. a_x(2) = 3$

$v_y(t) = t^3-3t$
$a_y(t) = dotv_y(t) =3t^2-3$
$:.a_y(2) = 9$

Hence, $|a| = sqrt(3^2+9^2)=sqrt90 = 3sqrt10$

And direction is given as:
$tantheta = 9/2$

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