Question

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

Answer 1

`a(t) = dv/dt = (2t, -pi sin pit - 3)` at t=2 one finds
`a(2) = (4,-3)` The magnitude f the acceleration vector is
`sqrt(16+9) = 5`
Its direction is given by an angle theta such that `tan theta = -3/4` or `theta = -37` degrees.