Particle \(P\) is moving anti-clockwise round a circle of radius \(\quantity{1}{m}\) at speed \(\quantity{1}{m\,s^{-1}}\).
The particle is at point \(A\) when \(t=0\) and its position at time \(\quantity{t}{s}\) is shown in the diagram below.
Combine some of the following diagrams to explain why the velocity vector \(\mathbf{v}\) is \((-\sin t, \cos t)\). You may be able to do this in more than one way. (The diagrams are also provided on cards.)
How have you used the fact that \(P\) is moving at unit speed on a unit circle?
What does this tell you about the derivatives of \(\cos t\) and \(\sin t\)?
Velocity is the rate of change of displacement, so its components are given by the rates at which its \(x\) and \(y\) coordinates are changing with respect to time. Therefore \[\mathbf{v}=\left( \dfrac{dx}{dt}, \dfrac{dy}{dt}\right).\]
