How to Find When an Object Returns from a Velocity-Time Graph

An ultrasonic motion sensor is fixed at one end of a straight track. A trolley starts 0.50 m from the sensor, rolls away, slows down, turns around, and travels back towards its starting point. At the turning point, its velocity is zero. Has it returned to where it started?

No. It has stopped momentarily, but it is still some distance from its starting position.

To find when an object returns to its starting position from a velocity-time graph, calculate the cumulative signed area under the graph. The object returns when the total signed area, measured from its starting time, first becomes zero.

This means the area above the time axis must balance the area below it. The important word is first: an object can return to its starting position more than once.

01Start with the physical situation

Imagine the trolley moving along a straight track.

• Moving away from the sensor is the positive direction.
• Moving towards the sensor is the negative direction.
• The trolley’s starting position is \(x_0=0.50\text{ m}\).

The sensor sends an ultrasonic pulse along the track. The pulse reflects from the trolley and travels back to the sensor. By measuring the time taken for the echo to return, the data logger calculates the trolley’s distance from the sensor.

This gives us position readings at different times. Software can then estimate velocity from how quickly the position changes.

What should the experiment show?

Suppose the trolley returns to its original position.

The sensor should record the same distance from the sensor that it recorded at the beginning:

\[ x=x_0=0.50\text{ m} \]

There is another way to identify this event. The trolley must have experienced zero net displacement since it started.

For a velocity-time graph, that means the positive and negative signed areas must cancel.

A simple stopwatch-and-marker method could give a rough return time, but human reaction time makes the measurement imprecise. It can also be difficult to identify repeated returns.

Using a motion sensor improves the method because it records position repeatedly throughout the motion. However, the sensor still takes readings at finite time intervals, and its velocity values are calculated estimates rather than perfectly precise instantaneous measurements.

Illustrative experimental data

Consider the following idealised readings from a trolley that starts 0.50 m from the sensor.

The trolley reaches its furthest position at \(t=3.0\text{ s}\). Its velocity is zero, but its position is 2.00 m, not 0.50 m.

It returns to its starting position at \(t=6.0\text{ s}\).

We can check the result using the areas under the velocity-time graph.

From 0 to 2 seconds, the positive rectangular area is:

\[ A_1=(2.0)(0.60)=+1.20\text{ m} \]

From 2 to 3 seconds, the positive triangular area is:

\[ A_2=\frac12(1.0)(0.60) =+0.30\text{ m} \]

From 3 to 4 seconds, the negative triangular area is:

\[ A_3=-\frac12(1.0)(0.60) =-0.30\text{ m} \]

From 4 to 6 seconds, the negative rectangular area is:

\[ A_4=(-0.60)(2.0) =-1.20\text{ m} \]

Adding the signed areas:

\[ \begin{aligned} \Delta x&=A_1+A_2+A_3+A_4\\ &=1.20+0.30-0.30-1.20\\ &=0\text{ m} \end{aligned} \]

The net displacement is zero, confirming that the trolley has returned to its starting position.

This is the central idea behind finding a return time from a velocity-time graph.

02The signed-area method

The area under a velocity-time graph represents displacement.

The sign of the area depends on which side of the horizontal axis it occupies.

• Above the time axis: positive displacement.
• Below the time axis: negative displacement.
• On the time axis: zero instantaneous velocity.

For an object moving along a straight line, the displacement is the sum of all signed areas.

\[ \Delta x=A_1+A_2+\cdots+A_n \]

Here, \(A_1,A_2,\ldots,A_n\) represent the signed areas of the individual sections of the graph.

The units also make sense:

\[ \text{Area}=\text{velocity}\times\text{time} \]

\[ =\text{m s}^{-1}\times\text{s} =\text{m} \]

Displacement is measured in metres.

The condition for returning to the starting position

An object returns to its starting position when:

\[ \boxed{\Delta x=0} \]

If the graph contains positive and negative regions, this is equivalent to:

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

where:

• \(A_+\) is the total positive area accumulated so far.
• \(A_-\) is the total negative signed area accumulated so far.

Notice the phrase accumulated so far. You must calculate the areas in chronological order if the question asks for the first return.

A reliable method for HSC exam questions

Use this sequence whenever you need to find when an object returns to its starting position from a velocity-time graph.

1. Check the axes. Confirm that the vertical axis represents signed velocity, not speed.
2. Choose the positive direction. Areas above the time axis are positive, and areas below are negative.
3. Divide the graph into sections. Split it at corners, changes in gradient, and crossings of the time axis.
4. Calculate each signed area. Use rectangles, triangles, or trapezia as appropriate.
5. Keep a running total. Add each new area to the previous cumulative displacement.
6. Find the first zero. The first time the cumulative displacement becomes zero is the first return to the starting position.

If the cumulative displacement does not become zero at a graph vertex, the object might return somewhere within the next section. You then need to calculate how much additional area is required and find the time taken to produce it.

This last step is where many otherwise correct solutions lose marks.

03Worked examples

04What if the graph is not velocity-time?

The signed-area method only applies directly when the graph’s vertical axis represents velocity.

If you are given a different type of graph, the method changes.

Position-time graphs

Suppose an object starts at position \(x=+3.0\text{ m}\).

It returns to its starting position when its position becomes \(+3.0\text{ m}\) again.

It does not need to reach \(x=0\).

This is a common source of mistakes. Students sometimes assume that returning to the starting position means reaching the coordinate origin. That is only true if the object originally started at the origin.

Displacement-time graphs

A displacement-time graph gives displacement relative to the starting position.

Therefore, the return condition is simply:

\[ \boxed{\Delta x=0} \]

You can read this directly from the vertical axis. There is no need to calculate an area.

Speed-time graphs

A speed-time graph is different because speed cannot be negative.

The graph tells you how fast an object moves, but not which direction it travels.

For example, an object could travel 4 m east and finish 4 m from its starting position. Alternatively, it could travel 2 m east and then 2 m west, returning to its starting position.

Both motions could have the same speed-time graph.

You need directional information to determine displacement and identify a return.

05Questions and solutions

The following questions progress from straightforward area calculations to situations where you must distinguish position, velocity, and cumulative displacement.

Question 1

A skateboarder starts at a marker and moves along a straight path.

The velocity is:

• \(+4.0\text{ m s}^{-1}\) from \(t=0\) to \(t=3.0\text{ s}\).
• \(-2.0\text{ m s}^{-1}\) after \(t=3.0\text{ s}\).

Assume both velocities remain constant during their respective intervals.

Calculate the time at which the skateboarder first returns to the starting position.

Solution 1

The skateboarder first returns to the starting position at \(t=9.0\text{ s}\).

First, calculate the positive displacement during the initial 3.0 seconds.

\[ \begin{aligned} A_+&=vt\\ &=(4.0)(3.0)\\ &=+12.0\text{ m} \end{aligned} \]

The skateboarder must travel 12.0 m in the negative direction to return.

Using the negative velocity:

\[ \begin{aligned} -12.0&=(-2.0)t\\ t&=6.0\text{ s} \end{aligned} \]

The return journey takes another 6.0 seconds.

Therefore:

\[ t_{\text{return}}=3.0+6.0 \]

\[ \boxed{t_{\text{return}}=9.0\text{ s}} \]

The skateboarder passes the starting marker after 9.0 seconds while travelling in the negative direction.

Question 2

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

• From \(t=0\) to \(t=2.0\text{ s}\), its velocity is constant at \(-3.0\text{ m s}^{-1}\).
• From \(t=2.0\) to \(t=6.0\text{ s}\), its velocity increases uniformly from \(-3.0\text{ m s}^{-1}\) to \(+5.0\text{ m s}^{-1}\).
• After \(t=6.0\text{ s}\), its velocity remains constant at \(+5.0\text{ m s}^{-1}\).

Determine the first time the trolley returns to its starting position.

Solution 2

The trolley first returns to its starting position at \(t=6.4\text{ s}\).

The important detail is that the trolley initially travels in the negative direction. We must accumulate negative displacement before adding the positive displacement.

Step 1: Calculate the initial negative displacement.

From 0 to 2.0 seconds, the graph forms a negative rectangle.

\[ \begin{aligned} A_1&=(-3.0)(2.0)\\ &=-6.0\text{ m} \end{aligned} \]

The trolley is now 6.0 m behind its starting position.

Step 2: Find when the velocity becomes zero.

Between 2.0 and 6.0 seconds, the velocity increases uniformly.

The acceleration is:

\[ \begin{aligned} a&=\frac{v-u}{t}\\ &=\frac{5.0-(-3.0)}{4.0}\\ &=2.0\text{ m s}^{-2} \end{aligned} \]

The velocity reaches zero after:

\[ 0=-3.0+(2.0)t \]

\[ t=1.5\text{ s} \]

This is 1.5 seconds after the start of the sloping section.

Therefore, the zero-velocity time is:

\[ t=2.0+1.5=3.5\text{ s} \]

Step 3: Calculate the negative triangular area.

From 2.0 to 3.5 seconds, the trolley still travels in the negative direction.

\[ \begin{aligned} A_2&=-\frac12(1.5)(3.0)\\ &=-2.25\text{ m} \end{aligned} \]

The cumulative displacement is now:

\[ \Delta x=-6.0-2.25 \]

\[ \Delta x=-8.25\text{ m} \]

The trolley is 8.25 m behind its starting position.

Step 4: Calculate the positive triangular area.

From 3.5 to 6.0 seconds, the velocity is positive.

\[ \begin{aligned} A_3&=\frac12(2.5)(5.0)\\ &=+6.25\text{ m} \end{aligned} \]

The cumulative displacement at 6.0 seconds is:

\[ \begin{aligned} \Delta x&=-6.0-2.25+6.25\\ &=-2.0\text{ m} \end{aligned} \]

The trolley is still 2.0 m behind its starting position.

Step 5: Calculate the remaining time.

After 6.0 seconds, the velocity is constant at \(+5.0\text{ m s}^{-1}\).

The trolley needs another \(+2.0\text{ m}\) of displacement.

\[ \begin{aligned} \Delta x&=vt\\ 2.0&=5.0t\\ t&=0.40\text{ s} \end{aligned} \]

Therefore:

\[ t_{\text{return}}=6.0+0.40 \]

\[ \boxed{t_{\text{return}}=6.4\text{ s}} \]

The trolley returns while travelling in the positive direction.

Notice that its velocity was zero at 3.5 seconds, but its cumulative displacement was \(-8.25\text{ m}\). That was its turning point, not its return time.

Question 3

An object moves along a straight track. Its position-time graph consists of straight sections joining the following points.

Determine:

a) The first time the object returns to its starting position.

b) The time at which the object passes through the coordinate origin, \(x=0\).

Solution 3

The object first returns to its starting position at \(t=8.0\text{ s}\). It passes through the coordinate origin at \(t=9.5\text{ s}\).

These are different events because the object starts at \(x=+3.0\text{ m}\), not at the origin.

Part a: Find the return time.

The initial position is:

\[ x_0=+3.0\text{ m} \]

For a position-time graph, the object returns when its position equals its initial position.

Looking at the supplied data, the position first returns to \(+3.0\text{ m}\) at:

\[ \boxed{t=8.0\text{ s}} \]

The object has returned to its starting position.

Part b: Find when the object passes through the origin.

From 8.0 to 10.0 seconds, the position changes linearly from \(+3.0\text{ m}\) to \(-1.0\text{ m}\).

The velocity during this section is:

\[ \begin{aligned} v&=\frac{\Delta x}{\Delta t}\\ &=\frac{-1.0-3.0}{10.0-8.0}\\ &=-2.0\text{ m s}^{-1} \end{aligned} \]

The object must travel 3.0 m in the negative direction to reach the origin.

\[ \begin{aligned} \Delta t&=\frac{3.0}{2.0}\\ &=1.5\text{ s} \end{aligned} \]

Therefore:

\[ t=8.0+1.5 \]

\[ \boxed{t=9.5\text{ s}} \]

The object returns to its starting position at 8.0 seconds, then continues moving and passes through the origin at 9.5 seconds.

This question tests an important distinction: returning to the starting position does not necessarily mean reaching zero position.

Question 4

A student is given a speed-time graph for an object moving along a straight track.

The graph shows:

• A speed of \(2.0\text{ m s}^{-1}\) at \(t=0\).
• A uniform decrease in speed to zero at \(t=2.0\text{ s}\).
• A uniform increase in speed to \(2.0\text{ m s}^{-1}\) at \(t=4.0\text{ s}\).

The student claims that the object must return to its starting position at \(t=4.0\text{ s}\), because it slows to zero and then moves again.

Assess this claim. Explain whether the graph provides enough information to determine the object’s final position. Support your explanation with two possible descriptions of the motion.

Solution 4

The student’s claim is not necessarily correct. A speed-time graph alone cannot determine whether the object returns to its starting position because it does not show the direction of motion.

The graph only gives the magnitude of velocity.

Step 1: Calculate the total distance travelled.

The graph contains two triangular regions.

The first triangle represents the distance travelled during the first two seconds:

\[ d_1=\frac12(2.0)(2.0) \]

\[ d_1=2.0\text{ m} \]

The second triangle represents the distance travelled during the next two seconds:

\[ d_2=\frac12(2.0)(2.0) \]

\[ d_2=2.0\text{ m} \]

The total distance is:

\[ d=2.0+2.0 \]

\[ \boxed{d=4.0\text{ m}} \]

This is all the graph can establish about the total distance.

Step 2: Consider the first possible motion.

The object moves east for the first two seconds, stops momentarily, and then continues east.

Its displacement is:

\[ \Delta x=+2.0+2.0 \]

\[ \Delta x=+4.0\text{ m} \]

It does not return to its starting position.

Step 3: Consider the second possible motion.

The object moves east for the first two seconds, stops momentarily, and then reverses direction.

It travels 2.0 m east and then 2.0 m west.

Its displacement is:

\[ \Delta x=+2.0-2.0 \]

\[ \Delta x=0\text{ m} \]

It returns to its starting position at 4.0 seconds.

Both motions produce exactly the same speed-time graph.

Therefore, the graph cannot distinguish between these possibilities.

The missing information is direction. A velocity-time graph would show the difference because velocity can be positive or negative.

Question 5

An object moves along a straight track. At \(t=0\), its position is \(x_0=+7.0\text{ m}\).

Its velocity-time graph passes through the following points. Straight lines connect consecutive points.

Determine every time between \(t=0\) and \(t=12\text{ s}\) when the object returns to its starting position.

At each return time, state its instantaneous velocity.

A student makes two claims:

• Claim A: The object only returns at \(t=12\text{ s}\), because the total positive area equals the total negative area over the entire graph.
• Claim B: The object returns at \(t=4\text{ s}\), \(t=9\text{ s}\), and \(t=11\text{ s}\), because these are the times when velocity becomes zero.

Assess both claims and explain the object’s motion.

Solution 5

The object returns to its starting position three times:

\[ \boxed{t=8.0\text{ s},\quad 10.0\text{ s},\quad 12.0\text{ s}} \]

Its velocity at these times is respectively:

\[ \boxed{-4.0,\quad +4.0,\quad -4.0\text{ m s}^{-1}} \]

Both student claims are incorrect.

The key is to calculate the cumulative displacement chronologically rather than looking only at the final area or the points where velocity becomes zero.

Step 1: Calculate the area from 0 to 2 seconds.

The graph forms a positive rectangle.

\[ A_1=(2)(4)=+8\text{ m} \]

The cumulative displacement is now \(+8\text{ m}\).

Step 2: Calculate the areas from 2 to 6 seconds.

The velocity decreases uniformly from \(+4\) to \(-4\text{ m s}^{-1}\).

It crosses the time axis halfway through this interval, at \(t=4.0\text{ s}\).

The positive triangular area is:

\[ A_2=\frac12(2)(4)=+4\text{ m} \]

The negative triangular area is:

\[ A_3=-\frac12(2)(4)=-4\text{ m} \]

The cumulative displacement at \(t=6.0\text{ s}\) is:

\[ \Delta x=8+4-4 \]

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

At \(t=4.0\text{ s}\), the cumulative displacement was \(+12\text{ m}\), even though velocity was zero.

The object has not returned.

Step 3: Calculate the area from 6 to 8 seconds.

The velocity is constant at \(-4\text{ m s}^{-1}\).

\[ A_4=(-4)(2) \]

\[ A_4=-8\text{ m} \]

Therefore:

\[ \Delta x=8-8 \]

\[ \boxed{\Delta x=0\text{ m}} \]

The object first returns to its starting position at:

\[ \boxed{t=8.0\text{ s}} \]

Its instantaneous velocity is still \(-4\text{ m s}^{-1}\).

Step 4: Calculate the areas from 8 to 10 seconds.

The velocity increases uniformly from \(-4\) to \(+4\text{ m s}^{-1}\).

It crosses zero at \(t=9.0\text{ s}\).

From 8 to 9 seconds, the negative triangular area is:

\[ A_5=-\frac12(1)(4) \]

\[ A_5=-2\text{ m} \]

The cumulative displacement becomes:

\[ \Delta x=0-2=-2\text{ m} \]

At \(t=9.0\text{ s}\), the object is 2.0 m behind its starting position. It has not returned, even though its velocity is zero.

From 9 to 10 seconds, the positive triangular area is:

\[ A_6=\frac12(1)(4) \]

\[ A_6=+2\text{ m} \]

Therefore:

\[ \Delta x=-2+2 \]

\[ \boxed{\Delta x=0\text{ m}} \]

The object returns to its starting position for the second time at:

\[ \boxed{t=10.0\text{ s}} \]

Its instantaneous velocity is \(+4.0\text{ m s}^{-1}\).

Step 5: Calculate the areas from 10 to 12 seconds.

The velocity decreases uniformly from \(+4\) to \(-4\text{ m s}^{-1}\).

It crosses zero at \(t=11.0\text{ s}\).

The positive triangular area from 10 to 11 seconds is:

\[ A_7=\frac12(1)(4) \]

\[ A_7=+2\text{ m} \]

The cumulative displacement becomes:

\[ \Delta x=0+2=+2\text{ m} \]

At \(t=11.0\text{ s}\), the object is 2.0 m ahead of its starting position. It has not returned.

The negative triangular area from 11 to 12 seconds is:

\[ A_8=-\frac12(1)(4) \]

\[ A_8=-2\text{ m} \]

The cumulative displacement becomes:

\[ \Delta x=2-2 \]

\[ \boxed{\Delta x=0\text{ m}} \]

The object returns for the third time at:

\[ \boxed{t=12.0\text{ s}} \]

Its instantaneous velocity is \(-4.0\text{ m s}^{-1}\).

Step 6: Assess the two claims.

Claim A is incorrect because the total positive and negative areas only establish that the net displacement is zero at \(t=12\text{ s}\). They do not establish that the object has not returned earlier.

The object has already returned at 8.0 seconds and 10.0 seconds.

Claim B is also incorrect because zero velocity means the object is momentarily stationary, not necessarily at its starting position.

At the three zero-velocity times, the object’s positions are:

None of these positions equals the initial position of \(+7.0\text{ m}\).

The complete set of return times is therefore:

The important distinction is that velocity tells you how the object is moving at an instant, while cumulative signed area tells you how far it has moved from its starting position.

06What to remember in an exam

When a question asks when an object returns to its starting position, first identify the graph type.

For a velocity-time graph, use cumulative signed area:

\[ \boxed{\text{Return occurs when }\Delta x=0} \]

For a position-time graph, find when the position equals its initial value.

For a displacement-time graph, find when displacement returns to zero.

Never assume that a zero-velocity crossing is a return, and never assume that the final time at which positive and negative areas balance is necessarily the first return.

Once you can identify the return time, the next useful step is to read the velocity at that instant. The area tells you where the object is; the height of the velocity-time graph tells you how it is moving when it gets there.

For a closer look at what happens when a velocity-time graph crosses or touches zero, see Can a Velocity-Time Graph Cross Zero While an Object Keeps Moving?.