From a Velocity Formula to a Motion Description
A velocity function gives more than the rate at which a particle’s position changes. Its sign tells you the particle’s direction of motion, and its zeros identify times when the particle is momentarily at rest. When you also differentiate velocity to find acceleration, the two signs together tell you whether speed is increasing or decreasing.
A complete analysis begins by finding the times when velocity or acceleration is zero, then testing signs on the intervals those times create. If the particle’s position is also supplied, its displacement over a time interval comes from subtracting the initial position from the final position. Displacement is signed: it records net change in position, not total distance traveled.
Reading Direction and Speed from Signs
To find a particle’s direction, solve \(v(t)=0\) and use those zeros to divide the time interval into pieces. On each piece, test the sign of \(v(t)\). The particle moves in the positive direction where \(v(t)>0\), and in the negative direction where \(v(t)<0\). At a zero of velocity, inspect the signs immediately before and after: a sign change indicates a change in direction, while no sign change means the particle is at rest at that instant without reversing.
Speed is the nonnegative quantity \(|v(t)|\). When \(v(t)\ne 0\), its rate of change is determined by both velocity and acceleration. If they have the same sign, speed increases; if they have opposite signs, speed decreases. Zeros of acceleration can mark boundaries where this pattern changes, so include them in the sign analysis.
Why the Speed Rule Works
Where \(v(t)>0\), speed is \(|v(t)|=v(t)\), so its derivative is \(a(t)\). Thus speed increases when \(a(t)>0\) and decreases when \(a(t)<0\). Where \(v(t)<0\), speed is \(|v(t)|=-v(t)\), so its derivative is \(-a(t)\). In that case, speed increases when \(a(t)<0\) and decreases when \(a(t)>0\). These two cases give the same-sign and opposite-sign rule.
This argument also explains why zeros of velocity need separate attention: the formula uses a fixed sign of \(v(t)\), so it applies only on intervals where velocity does not cross zero.
A Sign-Analysis Process
Solve \(v(t)=0\) within the time interval. These are candidate rest times and possible direction changes.
Differentiate \(v(t)\) to obtain \(a(t)\), then solve \(a(t)=0\). These times may separate speeding-up and slowing-down intervals.
Use a test time in every interval created by the zeros. Record the signs of both \(v\) and \(a\).
Use the sign of \(v\) for direction and the relationship between the signs of \(v\) and \(a\) for speed changes. Check either side of each velocity zero for a reversal.
Subtract the initial position from the final position and include the position unit.
Worked Example: Two Direction Changes and Several Speed Changes
Worked Example: Two Direction Changes and Several Speed Changes
A particle moves along a line for \(0\le t\le4\) seconds. Its velocity, in meters per second, is \(v(t)=(t-1)(t-3)\), and its position, in meters, is \(s(t)=\frac{t^3}{3}-2t^2+3t+2\). Find its direction and speed behavior, and calculate its displacement from \(t=0\) to \(t=4\).
Step 1: Find the velocity zeros. Solve \((t-1)(t-3)=0\), giving \(t=1\) and \(t=3\). The intervals for checking velocity signs are \((0,1)\), \((1,3)\), and \((3,4)\).
Step 2: Test the velocity signs. At \(t=0.5\), \(v(0.5)=(-0.5)(-2.5)=1.25>0\). At \(t=2\), \(v(2)=(1)(-1)=-1<0\). At \(t=3.5\), \(v(3.5)=(2.5)(0.5)=1.25>0\). The particle moves in the positive direction on \((0,1)\), the negative direction on \((1,3)\), and the positive direction on \((3,4)\). It changes direction at both velocity zeros.
Step 3: Find acceleration and its zero. Expanding the velocity gives \(v(t)=t^2-4t+3\), so \(a(t)=v'(t)=2t-4\). Acceleration is zero at \(t=2\); it is negative before \(2\) and positive after \(2\).
Step 4: Combine the signs. On \((0,1)\), \(v>0\) and \(a<0\), so speed decreases. On \((1,2)\), both are negative, so speed increases. On \((2,3)\), \(v<0\) and \(a>0\), so speed decreases. On \((3,4)\), both are positive, so speed increases.
Step 5: Calculate displacement from positions. Evaluate the supplied position function at both endpoints:
Therefore, \(\Delta s=s(4)-s(0)=\frac{10}{3}-2=\frac{4}{3}\text{ m}\). The positive sign means the final position is \(\frac{4}{3}\) meters in the positive direction from the initial position.
Check: Differentiating \(s(t)\) gives \(s'(t)=t^2-4t+3=(t-1)(t-3)=v(t)\), so the supplied position and velocity are consistent. The endpoint subtraction also gives \(s(4)-s(0)=\frac{4}{3}\text{ m}\), as stated.
Worked Example: Rest Without a Direction Change
Worked Example: Rest Without a Direction Change
For \(0\le t\le4\) seconds, let \(v(t)=(t-2)^2\) meters per second and \(s(t)=\frac{(t-2)^3}{3}+\frac{23}{3}\) meters. Determine the motion around \(t=2\), and find the displacement over the full interval.
Step 1: Analyze velocity. The only zero is \(t=2\). At \(t=1\), \(v(1)=(-1)^2=1>0\); at \(t=3\), \(v(3)=1^2=1>0\). The particle moves in the positive direction both before and after \(t=2\). It is at rest at \(t=2\), but it does not change direction.
Step 2: Analyze acceleration and speed. Differentiation gives \(a(t)=2(t-2)\), which is negative for \(t<2\) and positive for \(t>2\). Before \(t=2\), \(v>0\) and \(a<0\), so speed decreases. After \(t=2\), both are positive, so speed increases. The velocity is zero at \(t=2\), with positive velocity on both sides, so speed reaches zero there without a reversal.
Step 3: Find displacement. The endpoint positions are
Thus \(\Delta s=s(4)-s(0)=\frac{31}{3}-5=\frac{16}{3}\text{ m}\).
Check: The derivative of the position function is \(s'(t)=(t-2)^2=v(t)\). Subtracting the endpoint positions gives \(\frac{31}{3}-\frac{15}{3}=\frac{16}{3}\text{ m}\), confirming the displacement.
Worked Example: A Cubic Velocity and a Supplied Position
Worked Example: A Cubic Velocity and a Supplied Position
On \(0\le t\le2\) seconds, a particle has velocity \(v(t)=t(t^2-3)\) meters per second and position \(s(t)=\frac{t^4}{4}-\frac{3t^2}{2}+4\) meters. Find its direction and speed behavior, then calculate its displacement.
Step 1: Find the velocity zeros. Factoring \(v(t)=t(t^2-3)\) gives zeros \(t=0\) and \(t=\sqrt{3}\) on the interval. For \(0<t<\sqrt{3}\), \(t>0\) and \(t^2-3<0\), so \(v<0\). For \(\sqrt{3}<t\le2\), both factors are positive, so \(v>0\). The particle changes direction at \(t=\sqrt{3}\). At the initial endpoint \(t=0\), it is at rest.
Step 2: Find acceleration and test its sign. Expanding gives \(v(t)=t^3-3t\), so \(a(t)=3t^2-3=3(t-1)(t+1)\). On the given interval, acceleration is negative for \(0\le t<1\) and positive for \(1<t\le2\).
Step 3: Describe speed. On \((0,1)\), \(v<0\) and \(a<0\), so speed increases. On \((1,\sqrt{3})\), \(v<0\) and \(a>0\), so speed decreases. On \((\sqrt{3},2)\), \(v>0\) and \(a>0\), so speed increases.
Step 4: Calculate displacement from endpoint positions.
Therefore, \(\Delta s=s(2)-s(0)=2-4=-2\text{ m}\). The particle’s final position is two meters in the negative direction from its initial position.
Check: Differentiating the position gives \(s'(t)=t^3-3t=t(t^2-3)=v(t)\). The endpoint subtraction \(2\text{ m}-4\text{ m}=-2\text{ m}\) confirms the signed displacement.
Common Pitfall: Treating Every Rest as a Reversal
A zero of velocity means the particle is at rest at that instant, but it does not guarantee that the particle changes direction. For example, consider the incorrect claim that \(v(t)=(t-2)^2\) changes sign at \(t=2\) because it is zero there. The calculation fails: \(v(1)=1\) and \(v(3)=1\), so velocity is positive on both sides of \(2\). The particle stops momentarily and continues in the same direction.
A separate sign error can occur when finding displacement. If \(s(0)=4\) meters and \(s(2)=2\) meters, the incorrect subtraction \(4-2=2\) reverses the endpoints. The correct definition is final position minus initial position: \(\Delta s=s(2)-s(0)=2-4=-2\) meters.
Check Your Understanding
For each question, use signs and endpoint positions carefully. Include units where they are supplied.
- For \(v(t)=(t-1)(t-4)\), identify the rest times and determine the particle’s direction on \((0,1)\), \((1,4)\), and \((4,5)\).
- If \(v(t)=(t-3)^2\), is the particle’s direction the same on both sides of \(t=3\)? Explain using velocity signs.
- A particle has \(v(t)<0\) and \(a(t)<0\) on an interval. Is it speeding up or slowing down? State the sign relationship that supports your answer.
- If \(s(1)=7\) meters and \(s(5)=2\) meters, calculate the displacement from \(t=1\) to \(t=5\) and interpret its sign.
- For \(v(t)=t^2-4\) on \(0\le t\le3\), find the velocity zero and the acceleration zero, then describe the direction and speed behavior on the resulting intervals.
