How to Calculate Distance When a Velocity-Time Graph Crosses Zero

A trolley rolls away from an ultrasonic motion sensor, slows down, stops momentarily, and then rolls back towards the sensor. Its velocity-time graph starts above the time axis and crosses below it.

You calculate the area above the axis and get \(+12\text{ m}\). The area below the axis is \(-5\text{ m}\). Should the trolley’s total distance be \(12-5=7\text{ m}\), or \(12+5=17\text{ m}\)?

It is \(17\text{ m}\).

The crucial distinction is that displacement uses signed areas, while distance uses the magnitudes of the areas. When a velocity-time graph crosses the horizontal axis, you must separate the regions above and below the axis before calculating total distance.

01The rule: split at zero, then add the areas

First, imagine the trolley moving along a straight track.

• Above the time axis, its velocity is positive. It moves in the chosen positive direction.
• At the time axis, its velocity is zero. It is momentarily stationary.
• Below the time axis, its velocity is negative. It moves in the opposite direction.

When the graph crosses the axis, the trolley reverses direction.

Here is the important distinction:

The area below the axis is negative when calculating displacement. However, the trolley still travels a positive distance while moving in that direction.

Let:

• \(A_+\) = total area above the time axis.
• \(A_-\) = magnitude of the total area below the time axis.
• \(\Delta x\) = displacement.
• \(d\) = total distance travelled.

The two calculations are:

Displacement:

\[ \Delta x=A_+-A_- \]

Total distance:

\[ \boxed{d=A_++A_-} \]

Notice that both areas are added when finding distance.

For example, suppose the positive region has an area of \(20\text{ m}\), while the negative region has an area of \(8\text{ m}\).

The displacement is:

\[ \Delta x=20-8=+12\text{ m} \]

The distance travelled is:

\[ d=20+8=28\text{ m} \]

The object finishes \(12\text{ m}\) in the positive direction from where it started, despite travelling \(28\text{ m}\) altogether.

02Why does the area represent distance?

The area under a velocity-time graph represents displacement because:

\[ \Delta x=v\Delta t \]

Here, \(v\) is velocity in metres per second, and \(\Delta t\) is the time interval in seconds.

On a velocity-time graph, this multiplication is simply the area of a rectangle.

The units confirm the relationship:

\[ \text{Area}=(\text{m s}^{-1})(\text{s}) \]

\[ =\text{m} \]

So, the area represents a change in position.

When velocity is negative, the displacement contribution is negative because the object moves in the opposite direction. However, distance measures the total length of the path travelled, regardless of direction.

This is why we take the magnitude of each area when calculating distance.

03How to calculate distance when the graph crosses zero

Use this method whenever a velocity-time graph passes from positive to negative velocity, or vice versa.

Step 1: Identify every crossing of the time axis.

Each crossing indicates that velocity changes sign and the object reverses direction, assuming continuous one-dimensional motion.

Step 2: Find the time of each crossing.

If the graph contains a straight-line section, use the given points to calculate exactly when velocity reaches zero.

Step 3: Divide the graph into separate regions.

Treat each region above or below the time axis separately.

Step 4: Calculate the geometric area of each region.

Use the appropriate shape:

For a velocity-time graph, the horizontal dimension represents time and the vertical dimension represents velocity.

Step 5: Add all area magnitudes.

Do not subtract the areas below the axis.

\[ d=A_1+A_2+A_3+\cdots \]

If the question also asks for displacement, calculate the signed total separately.

Finding the exact zero-crossing time

A common mistake is to assume that a straight-line graph crosses the axis halfway between two plotted points.

That is only true when the two velocity values have equal magnitudes.

Consider a graph that changes from \(+6\text{ m s}^{-1}\) to \(-2\text{ m s}^{-1}\) over \(4\text{ s}\).

The velocity changes by \(8\text{ m s}^{-1}\) altogether. It takes longer to fall from \(+6\) to zero than from zero to \(-2\).

For a straight-line section joining two velocities of opposite signs, the crossing time is:

\[ t_0=t_1+ \frac{|v_1|}{|v_1|+|v_2|} (t_2-t_1) \]

Where:

• \(t_0\) is the time when velocity reaches zero.
• \(t_1\) and \(t_2\) are the initial and final times.
• \(v_1\) and \(v_2\) are the corresponding velocities.

This formula assumes a straight-line section between the two points.

For curved graphs, the crossing time must be determined from the curve or estimated using the available graph scale.

04Worked examples

05Common mistakes to avoid

There are a few particularly tempting errors in these questions.

Mistake 1: Subtracting the negative area when calculating distance.

This gives displacement, not distance.

Remember:

\[ \begin{aligned} \text{Distance}&=A_++A_-\\ \text{Displacement}&=A_+-A_- \end{aligned} \]

Mistake 2: Taking the absolute value of the final displacement.

Suppose the positive area is \(15\text{ m}\) and the negative area is \(10\text{ m}\).

Taking the absolute value of the signed total gives:

\[ |15-10|=5\text{ m} \]

That is the magnitude of displacement, not distance.

The actual distance is:

\[ 15+10=25\text{ m} \]

Mistake 3: Assuming the zero crossing occurs halfway through the interval.

A straight line does not necessarily cross the axis at its midpoint. Check the magnitudes of the endpoint velocities.

Mistake 4: Treating every negative velocity as negative distance.

Negative velocity indicates direction. It does not mean the object is travelling a negative distance.

Mistake 5: Forgetting additional crossings.

A graph can cross the time axis more than once. Every change of sign must be accounted for separately.

If the graph touches zero but returns to the same side, the object momentarily stops but does not reverse direction. Its area still contributes positively to distance.

One final limitation: if a question supplies only a few velocity measurements rather than a complete graph or an explicit interpolation rule, you cannot automatically assume that the sections between measurements are straight lines. The exact distance may not be uniquely determined.

06Questions and solutions

Try each question before reading its solution. The questions progress from a standard graph calculation to more demanding reasoning about what the graph actually allows you to conclude.

Question 1

A skateboarder moves along a straight path. Their velocity decreases uniformly from \(+10\text{ m s}^{-1}\) at \(t=0\text{ s}\) to \(-5\text{ m s}^{-1}\) at \(t=6\text{ s}\).

Calculate:

a) The time at which the skateboarder reverses direction.

b) The total distance travelled.

c) The displacement over the six-second interval.

Solution 1

The skateboarder travels a total distance of \(25\text{ m}\), with a displacement of \(+15\text{ m}\). The direction reversal occurs at \(t=4.0\text{ s}\).

Part a: Find the reversal time.

The graph changes from \(+10\text{ m s}^{-1}\) to \(-5\text{ m s}^{-1}\).

\[ \begin{aligned} t_0&=0+\frac{10}{10+5}(6)\\ &=\frac{60}{15}\\ &=\boxed{4.0\text{ s}} \end{aligned} \]

Part b: Calculate the total distance.

The positive region is a triangle with a base of \(4.0\text{ s}\) and a height of \(10\text{ m s}^{-1}\).

\[ A_+=\frac12(4)(10)=20\text{ m} \]

The negative region is another triangle, with a base of \(2.0\text{ s}\) and a height of \(5\text{ m s}^{-1}\).

\[ A_-=\frac12(2)(5)=5\text{ m} \]

Add the magnitudes:

\[ \begin{aligned} d&=20+5\\ &=\boxed{25\text{ m}} \end{aligned} \]

Part c: Calculate the displacement.

Subtract the negative area:

\[ \begin{aligned} \Delta x&=20-5\\ &=\boxed{+15\text{ m}} \end{aligned} \]

The skateboarder finishes \(15\text{ m}\) in the positive direction from the starting position.

Question 2

A motorised cart moves along a straight track. Its velocity-time graph consists of three sections:

• From \(t=0\text{ s}\) to \(t=2\text{ s}\), its velocity remains constant at \(+3\text{ m s}^{-1}\).
• From \(t=2\text{ s}\) to \(t=5\text{ s}\), its velocity decreases uniformly from \(+3\text{ m s}^{-1}\) to \(-6\text{ m s}^{-1}\).
• From \(t=5\text{ s}\) to \(t=7\text{ s}\), its velocity remains constant at \(-6\text{ m s}^{-1}\).

Calculate the total distance and displacement over the seven seconds. State when the cart reverses direction.

Solution 2

The cart travels \(25.5\text{ m}\), has a displacement of \(-10.5\text{ m}\), and reverses direction at \(t=3.0\text{ s}\).

Step 1: Find the zero crossing.

The velocity changes from \(+3\text{ m s}^{-1}\) to \(-6\text{ m s}^{-1}\) over three seconds.

\[ \begin{aligned} t_0&=2+\frac{3}{3+6}(5-2)\\ &=2+\frac{3}{9}(3)\\ &=\boxed{3.0\text{ s}} \end{aligned} \]

Step 2: Calculate the positive area.

There are two positive regions.

The first is a rectangle:

\[ A_1=3(2)=6\text{ m} \]

The second is a triangle between \(t=2\text{ s}\) and \(t=3\text{ s}\):

\[ A_2=\frac12(1)(3)=1.5\text{ m} \]

The total positive area is:

\[ A_+=6+1.5=7.5\text{ m} \]

Step 3: Calculate the negative area.

Again, there are two regions.

The first is a triangle between \(t=3\text{ s}\) and \(t=5\text{ s}\):

\[ A_3=\frac12(2)(6)=6\text{ m} \]

The second is a rectangle between \(t=5\text{ s}\) and \(t=7\text{ s}\):

\[ A_4=6(2)=12\text{ m} \]

Therefore:

\[ A_-=6+12=18\text{ m} \]

Step 4: Calculate the total distance.

\[ \begin{aligned} d&=A_++A_-\\ &=7.5+18\\ &=\boxed{25.5\text{ m}} \end{aligned} \]

Step 5: Calculate the displacement.

\[ \begin{aligned} \Delta x&=A_+-A_-\\ &=7.5-18\\ &=\boxed{-10.5\text{ m}} \end{aligned} \]

The negative displacement means the cart finishes \(10.5\text{ m}\) in the direction opposite to its original positive direction.

The key here is that the cart spends two seconds moving at a constant negative velocity after its reversal. That rectangle contributes to distance just as much as the triangular regions do.

Question 3

A trolley has a piecewise-linear velocity-time graph passing through the following points. Straight lines connect consecutive points.

Calculate:

a) Both times at which the trolley reverses direction.

b) The total distance travelled.

c) The net displacement.

d) The number of direction changes.

Solution 3

The trolley reverses direction twice, at \(t=\frac43\text{ s}\) and \(t=3.0\text{ s}\). Its total distance is \(\frac{28}{3}\text{ m}\), approximately \(9.33\text{ m}\), and its displacement is \(+6.0\text{ m}\).

Part a: Find both crossing times.

For the first section, velocity changes from \(+4\text{ m s}^{-1}\) to \(-2\text{ m s}^{-1}\).

\[ \begin{aligned} t_1&=0+\frac{4}{4+2}(2)\\ &=\boxed{\frac43\text{ s}} \end{aligned} \]

For the second section, velocity changes from \(-2\text{ m s}^{-1}\) to \(+2\text{ m s}^{-1}\).

The magnitudes are equal, so the crossing occurs halfway between \(2\text{ s}\) and \(4\text{ s}\).

\[ \boxed{t_2=3.0\text{ s}} \]

Part b: Calculate the total distance.

There are five regions to calculate.

The first positive triangle has a base of \(\frac43\text{ s}\) and a height of \(4\text{ m s}^{-1}\).

\[ A_1=\frac12\left(\frac43\right)(4) =\frac83\text{ m} \]

The first negative triangle has a base of:

\[ 2-\frac43=\frac23\text{ s} \]

Its area is:

\[ A_2=\frac12\left(\frac23\right)(2) =\frac23\text{ m} \]

The second negative triangle extends from \(t=2\text{ s}\) to \(t=3\text{ s}\).

\[ A_3=\frac12(1)(2)=1\text{ m} \]

The second positive triangle extends from \(t=3\text{ s}\) to \(t=4\text{ s}\).

\[ A_4=\frac12(1)(2)=1\text{ m} \]

Finally, the graph forms a positive rectangle from \(t=4\text{ s}\) to \(t=6\text{ s}\).

\[ A_5=2(2)=4\text{ m} \]

Add all five magnitudes:

\[ \begin{aligned} d&=\frac83+\frac23+1+1+4\\ &=\frac{10}{3}+6\\ &=\boxed{\frac{28}{3}\text{ m}}\\ &\approx\boxed{9.33\text{ m}} \end{aligned} \]

Part c: Calculate the displacement.

Now the negative areas must be subtracted.

\[ \begin{aligned} \Delta x &=\frac83-\frac23-1+1+4\\ &=2+4\\ &=\boxed{+6.0\text{ m}} \end{aligned} \]

Part d: State the number of direction changes.

There are two changes in the sign of velocity:

• At \(t=\frac43\text{ s}\), the trolley changes from positive to negative velocity.
• At \(t=3.0\text{ s}\), it changes from negative to positive velocity.

Therefore, it reverses direction twice.

Notice that the second crossing does not cancel the first crossing as an event. Both reversals contribute to the total distance travelled.

Question 4

A delivery robot travels along a straight path. Its velocity-time graph has one region above the time axis, followed by a triangular region below the axis.

The following information is known:

• Total distance travelled: \(50\text{ m}\).
• Net displacement: \(+14\text{ m}\).
• The negative triangular region has a base of \(6.0\text{ s}\).

Determine:

a) The total area above the time axis.

b) The magnitude of the area below the time axis.

c) The minimum velocity reached by the robot.

Explain why the robot’s distance is greater than the magnitude of its displacement.

Solution 4

The positive area is \(32\text{ m}\), the negative area has a magnitude of \(18\text{ m}\), and the minimum velocity is \(-6.0\text{ m s}^{-1}\).

Step 1: Set up the distance equation.

Distance is the sum of the two area magnitudes.

\[ A_++A_-=50 \]

Step 2: Set up the displacement equation.

The displacement is the positive area minus the magnitude of the negative area.

\[ A_+-A_-=14 \]

We now have two equations with two unknowns.

Step 3: Calculate the positive area.

Add the equations together:

\[ \begin{aligned} 2A_+&=64\\ A_+&=\boxed{32\text{ m}} \end{aligned} \]

Step 4: Calculate the negative area magnitude.

Substitute into the distance equation:

\[ \begin{aligned} 32+A_-&=50\\ A_-&=\boxed{18\text{ m}} \end{aligned} \]

Step 5: Calculate the minimum velocity.

The negative region is a triangle.

Its area is:

\[ A_-=\frac12 bh \]

Here, the base is \(6.0\text{ s}\), and the height is the magnitude of the minimum velocity.

Substitute:

\[ \begin{aligned} 18&=\frac12(6.0)|v_{\min}|\\ 18&=3|v_{\min}|\\ |v_{\min}|&=6.0\text{ m s}^{-1} \end{aligned} \]

Because the velocity is below the time axis, its sign is negative.

\[ \boxed{v_{\min}=-6.0\text{ m s}^{-1}} \]

The robot travels \(50\text{ m}\) because distance includes movement in both directions. However, its displacement is only \(+14\text{ m}\) because the return motion partially cancels the initial movement when calculating the change in position.

This question uses an important reverse technique: if you know both distance and displacement, you can work backwards to find the individual areas.

Question 5

A trolley moves along a straight track. A motion sensor provides the following information:

• At \(t=0\text{ s}\), its velocity is \(+6\text{ m s}^{-1}\).
• At \(t=4\text{ s}\), its velocity is \(-6\text{ m s}^{-1}\).
• Its net displacement over the four seconds is \(+12\text{ m}\).
• Its velocity changes continuously and its graph consists of straight-line sections, with exactly one crossing of the time axis.

No intermediate velocity measurements are available.

One student argues that the total distance must be \(12\text{ m}\), because the measured displacement is \(+12\text{ m}\).

Another student argues that the total distance cannot be determined uniquely because different velocity-time graphs could satisfy all the supplied information.

Which student is correct?

Demonstrate your reasoning by constructing two physically possible velocity-time graphs that satisfy every condition, but give different total distances.

Solution 5

The second student is correct. The total distance cannot be determined uniquely from the supplied information.

In fact, two valid velocity-time graphs can have the same initial velocity, final velocity, and displacement, but produce different total distances.

The key is understanding what the displacement tells us.

Let:

• \(A_+\) = total positive area.
• \(A_-\) = magnitude of the negative area.

We know:

\[ A_+-A_-=12 \]

However, distance is:

\[ d=A_++A_- \]

Knowing the difference between two areas does not tell us their sum.

Let’s construct two possible graphs.

Model A: The trolley maintains its initial velocity before slowing down.

Suppose the graph passes through these points:

Straight lines connect the points.

From 0 to 2 seconds, the graph forms a rectangle.

\[ A_1=6(2)=12\text{ m} \]

From 2 to 4 seconds, the graph decreases linearly from \(+6\) to \(-6\). It crosses the axis at \(t=3\text{ s}\).

The positive triangle has an area of:

\[ A_2=\frac12(1)(6)=3\text{ m} \]

The negative triangle has an area magnitude of:

\[ A_3=\frac12(1)(6)=3\text{ m} \]

Therefore, the total positive and negative areas are:

\[ A_+=12+3=15\text{ m} \]

\[ A_-=3\text{ m} \]

The displacement is:

\[ \Delta x=15-3=+12\text{ m} \]

The distance is:

\[ \begin{aligned} d_A&=15+3\\ &=\boxed{18\text{ m}} \end{aligned} \]

This graph satisfies all the supplied conditions.

Model B: The trolley initially increases its velocity before slowing down.

Now consider a different graph passing through these points:

Again, straight lines connect the points.

From 0 to 1 second, the graph forms a positive trapezium:

\[ \begin{aligned} A_1&=\frac12(6+15)(1)\\ &=10.5\text{ m} \end{aligned} \]

From 1 to 2 seconds, it forms another positive trapezium:

\[ \begin{aligned} A_2&=\frac12(15+0)(1)\\ &=7.5\text{ m} \end{aligned} \]

From 2 to 4 seconds, it forms a negative triangle:

\[ \begin{aligned} A_3&=\frac12(2)(6)\\ &=6\text{ m} \end{aligned} \]

The total positive area is:

\[ A_+=10.5+7.5=18\text{ m} \]

The magnitude of the negative area is:

\[ A_-=6\text{ m} \]

Therefore, the displacement is:

\[ \Delta x=18-6=+12\text{ m} \]

The distance is:

\[ \begin{aligned} d_B&=18+6\\ &=\boxed{24\text{ m}} \end{aligned} \]

Compare the two models.

Both graphs satisfy every condition in the question, yet their distances are different.

This proves that the exact distance cannot be determined from the supplied measurements alone.

The tempting mistake is to assume that knowing the displacement is enough to know the distance. It isn’t. Displacement fixes the difference between the positive and negative areas, but distance depends on their sum.

There is one useful conclusion we can still make. Since the graph has a genuine negative-velocity section, the trolley must have travelled further than the magnitude of its displacement:

\[ d>|\Delta x| \]

Therefore:

\[ d>12\text{ m} \]

But without more information about the shape of the graph, we cannot determine the exact value.

07What to remember for HSC Physics

When a velocity-time graph crosses the horizontal axis, your first job is to identify where the sign of velocity changes.

Then:

1. Find each zero crossing.
2. Split the graph into separate regions.
3. Calculate the geometric area of every region.
4. Add all area magnitudes to find total distance.
5. Use positive and negative signs to calculate displacement separately.

The distinction is simple, but it is easy to lose marks by mixing up the two quantities.

For a broader explanation of why velocity-time graph areas represent displacement, including the relationship between slope, area, and motion, see Velocity-Time Graphs: Slope, Area and Motion Explained.

If you’re unsure why a graph crossing zero represents a reversal of direction, read Can a Velocity-Time Graph Cross Zero While an Object Keeps Moving?.

The next useful distinction is between the area and gradient of a graph: area tells you displacement, while gradient tells you acceleration. Knowing which one the question is asking for is the first decision in almost every velocity-time graph problem.