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 ?

1 Answer
Dec 25, 2017

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

Explanation:

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