An object's two dimensional velocity is given by v(t) = ( t^2 +2t , cospit - 3t )v(t)=(t2+2t,cosπt3t). What is the object's rate and direction of acceleration at t=1 t=1?

1 Answer
Dec 25, 2017

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

Explanation:

a(t) =(dv(t))/(dt) =d/(dt)(t^2+2t, cos pit - 3t) = ( 2t+2, -pi sin pit-3)a(t)=dv(t)dt=ddt(t2+2t,cosπt3t)=(2t+2,πsinπt3)

a(1) = (4, -3)
The magnitude of the acceleration is
||a(1)|| = sqrt(4^2 + (-3)^2)= 5 ||a(1)||=42+(3)2=5

The direction of the acceleration is determined from

tantheta = (v_y)/(v_x) tanθ=vyvx
theta = tan^-1((v_y)/(v_x) )= tan^-1(-3/4)=-37^oθ=tan1(vyvx)=tan1(34)=37o
or

theta = 37^oθ=37o SE