When is the particle moving in the positive direction is a question that appears frequently in introductory calculus and physics courses. Understanding the answer requires linking the concepts of position, velocity, and the sign of the derivative that describes motion along a straight line. In this article we will explore the theoretical foundation, outline a clear step‑by‑step method for determining the intervals of positive motion, work through several illustrative examples, highlight typical pitfalls, and answer common questions that students encounter.
Understanding Motion Along a Line
A particle’s location on a number line is usually given by a function s(t), where t represents time and s denotes the position (measured in meters, feet, or any convenient unit). The instantaneous rate of change of position with respect to time is the velocity v(t), defined mathematically as the derivative
[ v(t)=\frac{ds}{dt}=s'(t). ]
Because velocity is a signed quantity, its sign tells us the direction of motion:
- v(t) > 0 → the particle is moving in the positive direction (to the right on a standard number line).
- v(t) < 0 → the particle is moving in the negative direction (to the left).
- v(t) = 0 → the particle is instantaneously at rest; it may be changing direction or pausing.
Thus, answering “when is the particle moving in the positive direction?” reduces to finding the time intervals where the velocity function is positive.
The Role of Velocity and Its Sign
Velocity can be constant, linear, quadratic, or any other functional form depending on the underlying forces acting on the particle. Regardless of its complexity, the sign analysis follows the same procedure:
- Find v(t) by differentiating s(t).
- Solve v(t) = 0 to locate critical points where the velocity changes sign.
- Test the sign of v(t) on each interval between consecutive critical points (including the intervals extending to −∞ and +∞ if the domain is unrestricted).
- Select the intervals where v(t) > 0; those are the times when the particle moves positively.
If the domain of t is restricted (for example, t ≥ 0 because negative time may not be physically meaningful), the analysis is performed only within that domain.
Step‑by‑Step Procedure
Below is a concise algorithm that can be applied to any position function s(t):
| Step | Action | Reason |
|---|---|---|
| 1 | Compute the derivative v(t) = s'(t). | Gives instantaneous velocity. |
| 2 | Set v(t) = 0 and solve for t. | Finds times where direction could change. |
| 3 | List all solutions in increasing order; add the domain boundaries if applicable. | Creates partition points for sign testing. |
| 4 | Choose a test point in each open interval and evaluate v(t) at that point. | Determines the sign of velocity on the interval. |
| 5 | Record intervals where v(t) > 0. | These are the answer intervals. |
| 6 | (Optional) Verify with a graph of v(t) or s(t) for visual confirmation. | Helps catch algebraic errors. |
Example Problems
Example 1: Quadratic Position
Suppose a particle moves along the x‑axis with position
[ s(t)=t^{3}-6t^{2}+9t+2,\qquad t\ge 0. ]
Step 1: Differentiate.
[v(t)=s'(t)=3t^{2}-12t+9. ]
Step 2: Solve v(t)=0.
[ 3t^{2}-12t+9=0;\Rightarrow;t^{2}-4t+3=0;\Rightarrow;(t-1)(t-3)=0, ]
so t = 1 and t = 3 seconds.
Step 3: Partition the domain [0,∞) using the critical points:
[ [0,1),;(1,3),;(3,\infty). ]
Step 4: Test each interval.
-
For t = 0.5 (in [0,1)):
[ v(0.5)=3(0.5)^{2}-12(0.5)+9=0.75-6+9=3.75>0. ] Positive → moving right. -
For t = 2 (in (1,3)):
[ v(2)=3(4)-24+9=12-24+9=-3<0. ]
Negative → moving left. -
For t = 4 (in (3,∞)):
[ v(4)=3(16)-48+9=48-48+9=9>0. ]
Positive → moving right.
Step 5: Collect positive intervals.
The particle moves in the positive direction for
[ 0\le t<1\quad\text{and}\quad t>3. ]
(At t = 1 and t = 3 the velocity is zero, so the particle is momentarily at rest.)
Example 2: Trigonometric Position
Consider [ s(t)=2\sin(t)-\frac{t}{2},\qquad 0\le t\le 4\pi. ]
Step 1: Velocity.
[ v(t)=s'(t)=2\cos(t)-\frac{1}{2}. ]
Step 2: Solve v(t)=0.
[ 2\cos(t)-\frac{1}{2}=0;\Rightarrow;\cos(t)=\frac{1}{4}. ]
The general solutions are
[ t=\arccos!\left(\frac{1}{4}\right)+2\pi k\quad\text{or}\quad t=-\arccos!\left(\frac{1}{4}\right)+2\pi k, ]
with k integer. Within [0,4π] we obtain four critical points:
[ t_{1}= \arccos!\left(\frac{1}{4}\right)\approx1.32,; t_{2}=2\pi-\arccos!\left(\frac{1}{4}\right)\approx4.96,; t_{3}=2\pi+\arccos!\left(\frac{1}{4}\right)\approx7.60,; t_{4}=4\pi-\arccos!\left(\frac{1}{4}\right)\approx11.24. ]
Step 3: Intervals:
[ [0,t_{1}),;(t_{1},t_{2}),;(t_{2},t_{3}),
