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

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Answer 1 · 2016-04-21T18:39:04

$a(3)=sqrt(3/2)$

$theta=54,74^o$

$a(t)=d/(d t) v(t)$

$a_x(t)=d/(d t)v_x(t)$

$a_x(t)=d/(d t)(sqrt(t-2))$

$a_x(t)=1/(2sqrt(t-1))" "a_x(3)=1/sqrt(3-1)" "a_x(3)=1/sqrt2$

$a_y(t)=d/(d t)v_y(t)$

$a_y(t)=d/(d t)(t^2-5t)$

$a_y(t)=2t-5" "a_y(3)=2*3-5" "a_y(3)=6-5$

$a_y(3)=1$

$a(3)=sqrt(a_x^2+a_y^2$

$a(3)=sqrt((1/sqrt2)^2+1^2)$

$a(3)=sqrt(1/2+1)$ $" ; "a(3)=sqrt(3/2)$

$tan theta=(a_y(3))/(a_x(3))$

$tan theta=1/(1/sqrt2)=sqrt2$

$theta=54,74^o$

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