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

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Answer 1 · 2016-04-21T13:20:41

$a(5)=12,37$
$theta=14,04 ^o$

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

$a_x(t)=d/(d t)(t^2+2t)=2t+2" "a_x(5)=2*5+2=12$

$a_y(t)=d/(d t)(cos pi t-3t)=-pi sinpit-3$

$a_y(5)=-pi sin5pi-3$

$sin 5 pi=0$

$a_y(5)=-3$

$a(5)=sqrt(a_x^2+a_y^2)$

$a(5)=sqrt(12^2+(-3)^2)$

$a(5)=sqrt(144+9)$

$a(5)=sqrt153$

$a(5)=12,37$

$tan theta=-3/12$

$tan theta=-0,25$

$theta=14,04 ^o$

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