Direction Is Determined by Velocity
A particle moving on a number line has a position coordinate that changes with time. Its velocity tells us how that coordinate is changing. If velocity is positive, position is increasing and the particle moves in the positive direction along the coordinate axis. If velocity is negative, position is decreasing and the particle moves in the negative direction.
The direction labels depend on the chosen coordinate axis. For example, if right is designated as positive, positive velocity means motion to the right. If left is designated as positive instead, the same physical motion has a positive velocity under that convention.
To describe direction over time, identify the intervals on which velocity is positive and the intervals on which it is negative. A time at which velocity is zero is a boundary between sign intervals, but the sign at that one time is not positive or negative. The values of velocity at the domain endpoints also matter: include an endpoint in a direction interval if velocity has the required sign there.
Why the Sign of Velocity Gives Direction
Velocity is the rate of change of position. Over a time interval where velocity stays positive, position must increase from the beginning of the interval to the end. The Mean Value Theorem makes this connection precise. If \(s\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(v(t)=s'(t)>0\) for every \(t\) in \((a,b)\), then for some \(c\) in \((a,b)\),
Here \(b-a>0\), and \(v(c)>0\), so \(s(b)>s(a)\). The position coordinate increases over the interval. If instead \(v(t)<0\) throughout \((a,b)\), the same theorem gives \(s(b)-s(a)<0\), so the position coordinate decreases. This justifies the direction interpretation: positive velocity corresponds to motion along the positive coordinate direction, and negative velocity corresponds to motion along the negative coordinate direction.
At a single time, the derivative describes the instantaneous rate. If \(v(t_0)>0\), then for sufficiently small positive time increments \(h\), the difference quotient \(\frac{s(t_0+h)-s(t_0)}{h}\) is positive, so \(s(t_0+h)>s(t_0)\). The particle’s position is increasing immediately after \(t_0\). The corresponding statement for \(v(t_0)<0\) gives a decrease. For interval descriptions, however, use the sign of velocity throughout the interval rather than inferring it from one sample value.
Finding Direction Intervals from a Formula
For a velocity formula, the basic task is to determine where \(v(t)>0\) and where \(v(t)<0\), within the stated time domain. Zeros of the formula divide the domain into intervals to test. On each interval, the sign is constant for a continuous formula that has no zero there. Then check the domain endpoints separately.
Worked Example: Direction from a Position Formula
A particle has position \(s(t)=t^2-6t+8\) meters for \(0\le t\le5\) seconds. Determine when it moves in each direction.
Differentiate position to obtain velocity:
Set the velocity equal to zero to find the sign boundary:
For \(0\le t<3\), \(2t-6<0\), so the particle moves in the negative direction. At \(t=3\), \(v(3)=2(3)-6=0\). For \(3<t\le5\), \(2t-6>0\), so the particle moves in the positive direction.
Check: At \(t=1\), \(v(1)=2(1)-6=-4\) m/s, confirming negative direction. At \(t=4\), \(v(4)=2(4)-6=2\) m/s, confirming positive direction. Differentiating the given position again gives \(s'(t)=2t-6\), the velocity used in the sign analysis.
Worked Example: Using Factors to Find Sign Intervals
Suppose \(v(t)=(t-1)(t-4)\) meters per second on \(0\le t\le5\) seconds. Determine the intervals of positive and negative direction.
The factors are zero at \(t=1\) and \(t=4\). These times divide the domain into \([0,1)\), \((1,4)\), and \((4,5]\), along with the two zero times.
On \(0\le t<1\), both factors are negative, so their product is positive. On \(1<t<4\), the first factor is positive and the second is negative, so their product is negative. On \(4<t\le5\), both factors are positive, so their product is positive. At \(t=1\) and \(t=4\), velocity is zero.
Check: These sample values confirm the signs in the three intervals: positive, negative, and positive. Substitution at the boundary times gives \(v(1)=(0)(-3)=0\) and \(v(4)=(3)(0)=0\). Thus the particle moves in the positive direction on \([0,1)\) and \((4,5]\), and in the negative direction on \((1,4)\).
Reading Direction from a Velocity-Time Graph
A velocity-time graph directly displays the sign of velocity. Where the graph is above the horizontal time axis, \(v(t)>0\), so the particle moves in the positive direction. Where it is below the axis, \(v(t)<0\), so the particle moves in the negative direction. Where it lies on the axis, \(v(t)=0\).
The height of the graph gives the value of velocity, not the particle’s position. To classify direction, compare the graph with the horizontal axis. Read the time intervals from the graph’s domain and mark every crossing or contact with the axis as a boundary to check.
Worked Example: Direction from a Piecewise Linear Velocity Graph
A velocity-time graph on \(0\le t\le6\) consists of line segments joining the points \((0,2)\), \((2,0)\), \((5,-3)\), and \((6,1)\), where time is measured in seconds and velocity in meters per second. Find the intervals of positive and negative direction.
On the first segment, the graph descends from \(v=2\) at \(t=0\) to \(v=0\) at \(t=2\). It is above the axis for \(0\le t<2\), and \(v(2)=0\).
The second segment runs from \((2,0)\) to \((5,-3)\). It is below the axis for \(2<t\le5\). On the final segment, velocity rises from \(-3\) at \(t=5\) to \(1\) at \(t=6\). Its slope is
Thus on that segment \(v(t)=-3+4(t-5)\). Find where it crosses the axis:
Since \(\frac{23}{4}=5.75\), the graph is below the axis for \(5<t<5.75\) and above it for \(5.75<t\le6\). Taken together, the particle moves in the positive direction on \([0,2)\) and \((5.75,6]\). It moves in the negative direction on \((2,5.75)\).
Check: At \(t=5.5\), the last segment gives \(v=-3+4(0.5)=-1\), which is negative. At \(t=6\), it gives \(v=-3+4(1)=1\), which is positive. At \(t=5.75\), it gives \(v=-3+4(0.75)=0\), confirming the crossing time. The first segment has slope \(\frac{0-2}{2-0}=-1\), so its equation is \(v=2-t\), which gives \(v(0)=2\) and \(v(2)=0\), as shown on the graph.
Direction Is Not Position or Speed
A particle’s position tells where it is relative to the origin; velocity tells which direction its position is changing. A positive position does not imply positive velocity. Likewise, speed is the magnitude of velocity and is nonnegative; it does not indicate direction on its own.
For example, let \(s(t)=10-t\) meters. At \(t=4\) seconds, the position is \(s(4)=10-4=6\) meters, which is positive. But the velocity is \(v(t)=s'(t)=-1\) meter per second, so the particle moves in the negative direction. The positive position means it is on the positive side of the origin, not that it is moving farther in that direction.
A Common Sign Mistake
A common error is to use the sign of position instead of the sign of velocity. In the example above, the incorrect calculation might be: “\(s(4)=6>0\), so the particle moves in the positive direction.” That conclusion does not follow. The correction is to differentiate position and inspect the velocity:
Therefore, at \(t=4\), the particle moves in the negative direction. Another frequent error is to assign a direction at a time when the velocity is zero. A zero velocity is neither positive nor negative; include such a time as a boundary rather than in either sign interval.
Check Your Understanding
For each question, use the sign of velocity to describe direction. Pay attention to the stated time domain and any times when velocity is zero.
- A particle has \(s(t)=3t-t^2\) for \(0\le t\le4\). Find the intervals of positive and negative direction.
- On \(0\le t\le5\), a velocity graph is above the time axis from \(t=0\) to \(t=2\), crosses the axis at \(t=2\), and remains below it until \(t=5\). State the direction intervals, including the status of the endpoints and the crossing time.
- If \(s(3)=-2\) meters and \(v(3)=4\) meters per second, what does each sign tell you?
- For \(v(t)=(t-2)(t-5)\) on \(0\le t\le6\), determine where the particle moves in the positive and negative directions.
- A particle has positive speed at an instant. Does this determine whether it moves in the positive or negative direction? Explain.
