An object's two dimensional velocity is given by $v(t) = ( 3t^2 - 5t , -t^3 +4t )$ . What is the object's rate and direction of acceleration at $t=1$ ?

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Answer 1 · 2016-05-04T17:59:44

$a(1)=sqrt2$
$alpha=45^o$

$a_x(t)=d/(d t) v_x(t)" ; "a_y(t)=d/(d t) v_y(t)$

$a_x(t)=d/(d t)(3t^2-5t)=6t-5$

$a_x(1)=6*1-5=1$

$a_y(t)=d/(d t) (-t^³+4t)=-3t^2+4$

$a_y(1)=-3*1^2+4$

$a_y(1)=-3+4$

$a_y(1)=1$

$a(1)=sqrt(a_x(1)^2+a_y(1)^2)$

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

$a(1)=sqrt2$

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$tan alpha=(a_y(1))/(a_x(1))=1/1=1" ; "alpha=45^o$

An object's two dimensional velocity is given by v(t) = ( 3t^2 - 5t , -t^3 +4t )v(t) = ( 3t^2 - 5t , -t^3 +4t ). What is the object's rate and direction of acceleration at t=1 t=1 ?