Reading Motion from a Velocity Graph
A velocity graph records how an object's velocity changes with time. Its vertical coordinate gives velocity, not position. Even so, the graph tells us when an object moves in either direction and how its position changes over an interval.
Two features do the main work: the graph's location relative to the horizontal axis indicates the direction of motion, and the signed area between the graph and the time axis gives displacement. A careful response connects those features to the requested conclusion rather than merely describing the shape of the graph.
Direction from the Sign of Velocity
On a velocity-versus-time graph, the horizontal axis represents \(v=0\). When the graph is above that axis, velocity is positive and the object moves in the positive direction. When the graph is below it, velocity is negative and the object moves in the negative direction. The positive direction is set by the context; for example, it may mean east, upward, or along a track from its starting point toward a marked endpoint.
A time at which the graph meets the axis has \(v=0\). That alone does not guarantee that the object changes direction. A direction change occurs when velocity changes sign: the graph must pass from above the axis to below it, or from below to above it. If it touches the axis and returns to the same side, the sign has not changed.
Signed Area and Change in Position
For an interval from \(t=a\) to \(t=b\), the displacement is the change in position, \(x(b)-x(a)\). The Fundamental Theorem of Calculus connects this change to velocity:
On a velocity graph, this definite integral is the signed area between the graph and the time axis. Regions above the axis contribute positive area; regions below the axis contribute negative area. Consequently, areas on opposite sides of the axis can cancel when finding net displacement.
The units provide a useful check. If velocity is measured in meters per second and time in seconds, area has units \((\text{meters per second})(\text{seconds})=\text{meters}\), which are position units. If an initial position is not specified, the graph still determines the displacement and the change in position, but not the object's absolute final position.
Why Signed Area Gives Displacement
Here is the connection behind the graph method. On a short time interval of length \(\Delta t\), if velocity is approximately constant at \(v\), the object's position changes by approximately \(v\Delta t\). This is a signed change: it is positive for positive velocity and negative for negative velocity. Adding these changes over smaller and smaller time intervals produces the definite integral of velocity.
More formally, since \(v(t)=x'(t)\), the Fundamental Theorem of Calculus gives:
The integral is represented geometrically by signed area. A thin region with width \(\Delta t\) and height near \(v(t)\) has area near \(v(t)\Delta t\), matching the approximate position change over that interval. Summing those signed areas gives the net change in position. This explains both why area is relevant and why area below the axis must count negatively.
Worked Example: A Graph That Crosses the Time Axis
Worked Example: A Graph That Crosses the Time Axis
A particle's velocity graph is horizontal at \(v=2\) meters per second from \(t=0\) to \(t=2\) seconds. From \(t=2\) to \(t=4\), it is a straight line from \((2,2)\) to \((4,-2)\). Describe the direction of motion and find the displacement from \(t=0\) to \(t=4\). Also state the position at \(t=3\) relative to the initial position.
Step 1: Find the zero of velocity on the sloping segment. Its slope is \(\frac{-2-2}{4-2}=-2\), so its equation is \(v(t)=2-2(t-2)=6-2t\). Setting \(v(t)=0\) gives \(6-2t=0\), hence \(t=3\).
Step 2: Describe direction. The graph is above the axis for \(0\le t<3\), so the particle moves in the positive direction during that time. It is below the axis for \(3<t\le4\), so the particle moves in the negative direction then. Velocity changes sign at \(t=3\), so the particle changes direction there.
Step 3: Add signed areas for displacement. From \(0\) to \(2\), the rectangle contributes \(2\cdot2=4\) meters. From \(2\) to \(3\), the triangle above the axis contributes \(\frac12(1)(2)=1\) meter. From \(3\) to \(4\), the triangle below the axis contributes \(-\frac12(1)(2)=-1\) meter. Thus the total displacement is \(4+1-1=4\) meters.
Step 4: Find the position change by \(t=3\). The displacement from \(0\) to \(3\) is \(4+1=5\) meters. Therefore, at \(t=3\), the particle is 5 meters in the positive direction from its initial position.
Check: On \(2\le t\le4\), integrating \(v(t)=6-2t\) gives \([6t-t^2]_2^4=(24-16)-(12-4)=0\). The two triangular areas on this segment cancel, leaving the total displacement of 4 meters from the first two seconds, as found by the graph areas.
Reading Position Changes Across Several Intervals
To describe position at multiple times, accumulate the signed areas in time order. The position at the end of an interval is the position at its beginning plus that interval's displacement. A negative displacement means position decreased; it does not by itself say that the final position is negative. Without a specified initial position, express the result relative to that initial position.
For example, if displacement on consecutive intervals is \(+3\) meters and then \(-5\) meters, the total displacement is \(-2\) meters. The object ends 2 meters in the negative direction from where it started, regardless of its absolute starting coordinate.
Worked Example: Combining Areas Below and Above the Axis
A particle's velocity graph is a line segment from \((0,-3)\) to \((2,-3)\), followed by a straight segment through \((5,0)\) and \((7,2)\). Time is measured in seconds and velocity in meters per second. Find the displacement from \(t=0\) to \(t=7\), and describe how the position changes.
Step 1: Read direction from the signs. The graph is below the axis for \(0\le t<5\), so the particle moves in the negative direction. It is above the axis for \(5<t\le7\), so it moves in the positive direction. The sign changes at \(t=5\), so the particle changes direction then.
Step 2: Find the signed areas. From \(0\) to \(2\), the rectangle is below the axis, so its signed area is \(-(2)(3)=-6\) meters. From \(2\) to \(5\), the triangle has base \(3\) seconds and height \(3\) meters per second, so its signed area is \(-\frac12(3)(3)=-\frac92\) meters. From \(5\) to \(7\), the triangle above the axis has signed area \(\frac12(2)(2)=2\) meters.
Step 3: Combine the contributions. The total displacement is \(-6-\frac92+2=-\frac{12}{2}-\frac92+\frac42=-\frac{17}{2}\) meters, or \(-8.5\) meters. The particle's final position is therefore 8.5 meters in the negative direction from its initial position.
Check: The sloping segments both lie on \(v(t)=t-5\). Their combined integral from \(2\) to \(7\) is \(\left[\frac{t^2}{2}-5t\right]_2^7=(24.5-35)-(2-10)=-10.5+8=-2.5\) meters. Adding the first-interval displacement of \(-6\) meters gives \(-8.5\) meters, agreeing with the signed-area sum.
When an Initial Position Is Known
If a problem gives an initial position, use displacement to find the final position: final position equals initial position plus displacement. The graph determines the change; the initial value anchors that change to an absolute position.
Worked Example: Using a Curved Velocity Graph
A particle starts at position \(x(0)=10\) meters. Its velocity is \(v(t)=t^2-4\) meters per second for \(0\le t\le3\). Use the graph's sign and signed area to describe its direction and find its position at \(t=3\).
Step 1: Determine where velocity is zero. Solve \(t^2-4=0\), so \(t=-2\) or \(t=2\). Only \(t=2\) lies in the stated time interval. For \(0\le t<2\), \(t^2-4<0\); for \(2<t\le3\), \(t^2-4>0\). Thus the particle moves in the negative direction until \(t=2\), then in the positive direction.
Step 2: Find the total signed area. Displacement is \(\int_0^3(t^2-4)\,dt=\left[\frac{t^3}{3}-4t\right]_0^3=(9-12)-0=-3\) meters. The net displacement is negative even though the particle moves positively during the last second.
Step 3: Use the initial position. The final position is \(x(3)=x(0)+\int_0^3v(t)\,dt=10+(-3)=7\) meters.
Check: The antiderivative \(\frac{t^3}{3}-4t\) differentiates to \(t^2-4\), the given velocity. Its change from 0 to 3 is \(-3\), so adding that displacement to the initial position gives \(7\) meters.
A Common Mistake: Treating Every Area as Positive
A frequent error is to add the magnitudes of all regions between the graph and the axis when the question asks for displacement. That procedure ignores the direction encoded by the sign of velocity. In the preceding linear-graph example, adding magnitudes gives \(6+4.5+2=12.5\) meters. This is not the displacement: it counts the below-axis regions as positive.
The correction is to attach a sign to each area before adding. The first two regions are below the axis and contribute \(-6\) and \(-4.5\) meters; the final region is above it and contributes \(+2\) meters. Therefore displacement is \(-6-4.5+2=-8.5\) meters. The sign says the final position is lower than the initial position by 8.5 meters.
Check Your Understanding
Use the signs of velocity and the signed areas to answer each question.
- A velocity graph is above the time axis from \(t=1\) to \(t=4\). In which direction does the particle move on that interval?
- A graph meets the time axis at \(t=3\) but remains below it on both sides of \(t=3\). Does the particle change direction at that time? Explain.
- A region above the axis has area \(7\) meters, and a later region below the axis has area magnitude \(10\) meters. What is the combined displacement?
- A particle begins at position \(x(0)=4\) meters and has displacement \(-6\) meters over an interval. What is its final position?
- Explain why a velocity graph's displacement is not found by adding the magnitudes of all regions between the graph and the time axis.
