Question
At time t, the position of a particle moving in the xy-plane is given by the parametric functions (x(t), y(t)), where \(\frac{dx}{dt}=t^{2}+sin(3t^{2}).\) The graph of y, consisting of three line segments, is shown in the figure above. At t = 0, the particle is at position (5, 1).
(a) Find the position of the particle at t = 3.
(b) Find the slope of the line tangent to the path of the particle at t = 3.
(c) Find the speed of the particle at t 3.
(d) Find the total distance traveled by the particle from t = 0 to t = 2.
Answer/Explanation
Ans:
(a) v(4) = 2.978716 > 0
v'(4) = -1.164000 < 0
The particle is slowing down since the velocity and acceleration have different signs.
(b) v(t) = 0 ⇒ t = 2.707468
v(t) changes from positive to negative at t = 2.707. Therefore, the particle changes direction at this time.
(c) \(x(0)=x(4)+\int_{4}^{0}v(t)dt\)
= 2 + (-5.815027) = -3.815
(d) \(Distance=\int_{0}^{3}v|(t)|dt=5.301\)