V(1/3) = (2π(1/3), 2π(1/3) = (2π/3 , 2π/3)
let the angle of the vector at that moment be Ø
tanØ = (2π/3)/(2π/3) = 1
"after the next 1/3 sec", implies t = 2/3
V(2/3) = (4π/3 ,4π/3)
since tanØ = (4π/3)/(4π/3) = 1
the angle has not changed, so the angular velocity is 0
Let Vector V(t) = 2pi t î + 2pi t ĵ.
Find the vector V(1/3).
Where is the vector after the next 1/3s?
What is the angular speed?
1 answer