An object's two dimensional velocity is given by #v(t) = ( sqrt(t-2)-2t , t^2)#. What is the object's rate and direction of acceleration at #t=3 #?
2nd quandrant (negative x and positive y)
When you put 3 (for t), you will get v(t) = (-5, 9). The distance (if speed is expressed as m/s and time is expressed as s) 30.89 meters.
Note that the units of time and velocity were not provided in the original question.
The acceleration is the derivative of the velocity
The rate of acceleration is
The angle is in the 2nd quadrant and is