- Therefore, the particle travels a distance of 48.
- Velocity is 0; acceleration is 0.
This should not be a surprise because 2sin^2 t + 2cos^2 t = 2, so the position is a constant. This
means that the particle is not moving and thus has a velocity and acceleration of 0.
- t = ≈ 0.122
In order to find where the particle is changing direction, we need to find where the velocity of
the particle changes sign. The velocity function of the particle is the derivative of the position
function: v(t) = 3t 2 + 16t − 2. Next, we set the velocity equal to zero. The solutions are t =
and t = . We can eliminate the second solution because it is negative
and we are restricted to t > 0. Next, we check the sign of the velocity on the intervals 0 < t <
and t > . When 0 < t < , the velocity is negative, so the
particle is moving to the left. When t > , the velocity is positive, so the particle is
moving to the right. Therefore, the particle is changing direction at t = .
- The velocity is never 0, which means that it never changes signs and thus the particle does not
change direction.
In order to find where the particle is changing direction, we need to find where the velocity of
the particle changes signs. The velocity function of the particle is the derivative of the position
function: = v(t) = 6t^2 − 12t + 12. Next, we set the velocity equal to zero. There are no real
solutions. If we try a few values, we can see that the velocity is always positive. Therefore, the
particle does not change direction.
SOLUTIONS TO PRACTICE PROBLEM SET 14
