An object's two dimensional velocity is given by $v(t) = ( sqrt(t-2)-2t , t^2)$ . What is the object's rate and direction of acceleration at $t=7$ ?

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Answer 1 · 2016-05-21T10:04:54

$a(7)=14.11" "m/s^2$
$tan alpha=-7.87$

$v(t)=(sqrt(t-2)-2t,t^2)$

$v_x(t)=sqrt(t-2)-2t$

$v_y(t)=t^2$

$a_x(t)=d/(d t) v(t)=d/(d t)(sqrt(t-2)-2t)=1/(2*sqrt(t-2))-2$

$a_x(7)=1/(2sqrt(7-2))-2=1/(2sqrt5)-2=(1-2(2sqrt5))/(2sqrt5)$

$a_x(7)=(1-4sqrt5)/(2sqrt5)=(2sqrt5-40)/20=(sqrt5-20)/10$

$a_x(7)=-1.78$

$a_y(t)=d/(d t)v_y(t)=d/(d t)(t^2)=2t$

$a_y(7)=2*7=14$

$a(7)=sqrt(a_x^2+a_y^2)$

$a(7)=sqrt((-1*78)^2+14^2$

$a(7)=14.11" "m/s^2$

$tan alpha=a_y/a_x$

$tan alpha=-14/(1.78)$

$tan alpha=-7.87$

An object's two dimensional velocity is given by v(t) = ( sqrt(t-2)-2t , t^2)v(t) = ( sqrt(t-2)-2t , t^2). What is the object's rate and direction of acceleration at t=7 t=7 ?