An object's two dimensional velocity is given by $v(t) = ( sqrt(3t)-t , t^2-5t)$ . What is the object's rate and direction of acceleration at $t=2$ ?

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Answer 1 · 2017-04-14T09:31:29

The rate of acceleration is $=1.07ms^-2$ in the direction $=248.7º$

The acceleration is the derivative of the velocity.

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

$v(t) = < sqrt(3t)-t , t^2-5t>$

$a(t)= < sqrt3*1/(2sqrtt)-1,2t-5>$

When $t=2$

$a(2) = < sqrt3*1/(2sqrt2)-1,2*2-5 >$

$=< -0.39,-1 >$

The rate of acceleration is

$=||a(2)||$

$=||< -0.39,-1 >||$

$=sqrt((0.39)^2+(1)^2)$

$=1.07ms^-2$

The direction is in the 3rd quadrant

$theta=180+arctan(1/0.39)$

$=248.7º$

An object's two dimensional velocity is given by v(t) = ( sqrt(3t)-t , t^2-5t)v(t) = ( sqrt(3t)-t , t^2-5t). What is the object's rate and direction of acceleration at t=2 t=2 ?