An object's two dimensional velocity is given by $v(t) = (3t-sqrtt , 2t-4 )$ . What is the object's rate and direction of acceleration at $t=a$ ?

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Answer 1 · 2017-03-12T19:04:28

The rate of acceleration is $=sqrt(13-3/sqrta+1/(4a))$ in the direction $=arctan((4sqrta)/(6sqrta-1))$

$v(t)=(3t-sqrtt,2t-4)$

The acceleration is the derivative of the velocity

$v'(t)=(3-1/(2sqrtt),2)$

Therefore,

$v'(a)=(3-1/(2sqrta),2)$

The rate of acceleration is

$||v'(a)||=sqrt((3-1/(2sqrta))^2+ 4)$

$=sqrt(9-3/sqrta+1/(4a)+4)$

$=sqrt(13-3/sqrta+1/(4a))$

The direction is

$=arctan(2/(3-1/(2sqrta)))$

$=arctan((4sqrta)/(6sqrta-1))$

An object's two dimensional velocity is given by v(t) = (3t-sqrtt , 2t-4 )v(t) = (3t-sqrtt , 2t-4 ). What is the object's rate and direction of acceleration at t=a t=a ?