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 $t=3$ ?

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Answer 1 · 2017-06-11T05:31:38

The rate of acceleration is $=3.14ms^-2$ in the direction $=39.6º$ anticlockwise from the x-axis.

The acceleration is the derivative of the velocity

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

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

So,

when $t=3$

$a(3)=(3-1/sqrt3,2)=(2.42,2)$

The rate of acceleration is

$=||a(3)||=sqrt(2.42^2+(2)^2)=sqrt9.87=3.14$

The direction is $arctan(2/2.42)=39.6º$

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=3 t=3 ?