The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. What is the angle?

θ = 90º

Explanation:

The velocity is given by

          v =

calculate

          v = 3 i ^ + √2 j ^ + 2t k ^

acceleration is defined by

         a = dv / dt

         a = 2 k ^

one way to find the angle is with the dot product

         v. a = | v | | a | cos θ

         cos θ= v.a / | v | | a |

Let's look for the value of each term

       v. a = 4 t

        | v | = =

        | a | = 2

they ask us for the angle for time t = 0

         v. a = 0

        | v | = √11 = 3.317

we substitute

        cos θ = 0 /√11

        cos θ = 0

therefore the angles must be θ = 90º

d eras and periods

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