An object's two dimensional velocity is given by #v(t) = ( t-2t^3 , 4t^2-3t)#. What is the object's rate and direction of acceleration at #t=3 #?
The acceleration is the first derivative of the velocity,
which, evaluated at t=3 in standard form is
Taking the first derivative of the velocity, we get
Evaluating this at
The magnitude of this acceleration is found by Pythagorus' theorem because the two components above are perpendicular:
The direction is found from
But this angle cannot be right, because it would describe an acceleration vector pointing into quadrant 4 - with positive
So, you must be careful in how you interpret this last result. If you did the
A diagram will sort out the problem. With a negative
Always include a diagram when solving two-dimensional vector problems. It will help you avoid errors in calculation.