Question 3.3: The position of a particle moving along the x-axis varies wi......
The position of a particle moving along the x-axis varies with time t according to the relation x = t³ − 12 t + 20, where x is given in meters and t in seconds. (a) Find the velocity and the acceleration of the particle as a function of time. (b) Is there ever a time when v = 0? (c) Describe the particle’s motion for t ≥ 0.
(a) To get the velocity v as a function of time t, we differentiate the coordinate x with respect to t as follows:
v=\frac{d x}{d t}=\frac{d}{d t}\left(t^{3}-12t+20\right)~~\Rightarrow~~v=3\,t^{2}-12To get the acceleration a as a function of time t, we differentiate the velocity v with respect to t as follows:
a=\frac{d v}{d t}=\frac{d}{d t}\left(3\,{t}^{2}-12\right)\,\,\,\,\Rightarrow\,\,\,\,\,a=6\,{t}(b) Setting v = 0 in the velocity relation yields:
0=3\,t^{2}-12.which has the solution t = ±2 s. The negative answer has to be rejected, since time must be always positive. Thus at t = 2 s the velocity of the particle is zero.
(c) To describe the particle’s motion for t ≥ 0 we examine the expressions
x = t³ − 12 t + 20, v = 3 t² − 12, and a = 6 t.
At t = 0, the particle is at x = 20 m from the origin and moving to the left with velocity v = −12 m/s and not accelerating since a = 0, see Fig.3.9.
At 0 < t < 2 s, the particle continues to move to the left (x decreases), but at a decreasing speed, because it is now accelerating to the right, a = positive (Check the expressions of x, v, and a for t = 1 s and compare the results with Fig. 3.9).
At t = 2 s, the particle stops momentarily (v = 0) to reverse its direction of motion. At this moment x = 4 m, i.e. it will be as close as it will ever be to the origin. It will continue to accelerate to the right at an increasing rate, see Fig. 3.9.
For t > 2 s, the particle continues to accelerate and move to the right, and its velocity, which is now to the right, increases rapidly, see Fig. 3.9.
