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 $t=1$ ?

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Answer 1 · 2017-12-25T19:44:27

$a = 5$
$theta = - 37^o " or " 37"^o$ SE

$a(t) =(dv(t))/(dt) =d/(dt)(t^2+2t, cos pit - 3t) = ( 2t+2, -pi sin pit-3)$

a(1) = (4, -3)
The magnitude of the acceleration is
$||a(1)|| = sqrt(4^2 + (-3)^2)= 5$

The direction of the acceleration is determined from

$tantheta = (v_y)/(v_x)$
$theta = tan^-1((v_y)/(v_x) )= tan^-1(-3/4)=-37^o$
or

$theta = 37^o$ SE

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=1 t=1 ?