Can Two Objects Meet with Different Position-Time Slopes?

Two carts roll towards each other along a straight track. The left cart moves to the right, while the right cart moves to the left. On a position-time graph, one line slopes upwards and the other slopes downwards.

Can they still meet, even though their graphs have different slopes?

Yes. Two objects meet when they have the same position at the same instant. Their slopes do not need to be equal. The slope tells you their velocity, not their position.

In fact, different slopes are often exactly what allow two objects to approach each other and meet.

Before going further, consider this: if two lines cross on a position-time graph, what must be true at the crossing? Both objects must have the same position at the same time. Their velocities can be completely different.

01What does an intersection actually mean?

A position-time graph has:

• Time, \(t\), on the horizontal axis.
• Position, \(x\), on the vertical axis.

Every point on the graph tells you where an object is at a particular time.

Suppose two objects have position-time relationships \(x_A(t)\) and \(x_B(t)\).

They meet when:

\[ \boxed{x_A(t)=x_B(t)} \]

This is the condition you need to remember.

Notice what is missing from the equation: velocity.

There is no requirement for the objects to have equal velocities when they meet.

What does the slope tell us?

The slope of a position-time graph gives velocity:

\[ v=\frac{\Delta x}{\Delta t} \]

Here:

• \(v\) is velocity in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(\Delta x\) is the change in position in metres (m).
• \(\Delta t\) is the time interval in seconds (s).

For a straight line, the slope is constant, so the object has constant velocity.

For a curved graph, the gradient of the tangent at a particular point gives the instantaneous velocity.

A positive slope means motion in the positive direction. A negative slope means motion in the negative direction.

A steeper slope means a greater speed, provided both graphs use the same axis scales.

Consider these three situations:

There is an important distinction here.

Two straight lines with different slopes will intersect somewhere if extended indefinitely. However, that intersection might occur before the experiment starts or after the period being investigated.

You must check that the intersection occurs at a physically relevant time.

The mathematics behind the intersection

For constant velocity, position can be calculated using:

\[ x=x_0+vt \]

where:

• \(x\) is the position at time \(t\).
• \(x_0\) is the initial position.
• \(v\) is the constant velocity.
• \(t\) is the elapsed time.

For two objects:

\[ x_A=x_{A0}+v_At \]

\[ x_B=x_{B0}+v_Bt \]

At their meeting point:

\[ x_{A0}+v_At=x_{B0}+v_Bt \]

Rearranging gives:

\[ \boxed{t_m=\frac{x_{B0}-x_{A0}}{v_A-v_B}} \]

This equation gives the time at which their positions are equal, assuming constant velocities and a common starting time.

The denominator is the difference between their velocities. It describes how quickly their separation changes.

If both objects have the same velocity but start at different positions, they remain separated. Their graphs are parallel.

If they start at the same position and have the same velocity, they remain together.

02How an experiment would test this

It is useful to understand how the graph would actually be produced.

Imagine investigating two small carts moving towards each other along a low-friction track.

The question is whether their different velocities prevent them from meeting.

Setting up the experiment

Place a metre ruler parallel to the track, with its zero mark at the left-hand end.

Position two carts facing each other. Place an overhead camera so that both carts and the ruler are visible in the same recording.

Start recording, then release the carts.

For each video frame, record the positions of the two facing edges of the carts. Using the same video means both positions correspond to the same instant.

This last detail is important. You cannot reliably compare two position-time graphs if their time origins or position origins are inconsistent.

What should the experiment show?

If the two carts have different constant velocities, their position-time graphs should have different slopes.

If their facing edges reach the same position at the same time, the graphs should intersect.

The following is an illustrative data set, not measured experimental results.

Look at the changes.

Cart A moves \(0.40\ \mathrm{m}\) to the right every second.

Cart B moves \(0.20\ \mathrm{m}\) to the left every second.

Their position-time relationships are:

\[ x_A=0.30+0.40t \]

\[ x_B=2.10-0.20t \]

At \(t=3.0\ \mathrm{s}\), both facing edges are at \(x=1.50\ \mathrm{m}\).

The experiment’s predicted meeting point is therefore:

\[ (t,x)=(3.0\ \mathrm{s},1.50\ \mathrm{m}) \]

The slopes are still different at that point.

Why does the experimental design matter?

Suppose you filmed cart A using one phone and cart B using another.

If the clocks started at different times, the two position-time graphs would not share the same time origin. Their apparent intersection could be misleading.

Similarly, using different position references could shift one graph vertically.

Using one camera and one ruler controls these problems. If separate cameras are necessary, their clocks must be synchronised and their position measurements calibrated to a common origin.

Real measurements also have uncertainty. A limited video frame rate might mean that the exact instant of contact falls between two recorded frames. A line of best fit can help estimate the meeting time.

The important result is that different gradients do not prevent the two position traces from reaching the same coordinate simultaneously.

04Using signed areas on velocity-time graphs

So far, we have used position-time graphs to find when objects meet.

But what if a question gives you velocity-time graphs instead?

You can still determine whether the objects meet. You just need one additional idea: the signed area under a velocity-time graph represents displacement.

Displacement is the change in position, including direction.

For example, if right is positive:

• An area above the time axis represents positive displacement.
• An area below the time axis represents negative displacement.

The area is signed because direction matters.

Connecting displacement to position

Suppose an object starts at position \(x_0\).

Its position at a later time is:

\[ \boxed{x=x_0+\Delta x} \]

where \(\Delta x\) is the signed area under its velocity-time graph over the relevant time interval.

For two objects to meet:

\[ x_{A0}+\Delta x_A = x_{B0}+\Delta x_B \]

This equation is especially useful when the velocity changes during the motion.

You can calculate the signed areas using familiar shapes:

Remember that areas below the time axis must be treated as negative.

A common mistake: Using total distance instead of displacement

In the previous example, object B always moved left. Its negative signed areas correctly represented its displacement.

However, an object can change direction during a question.

Suppose a velocity-time graph has a section below the time axis followed by a section above it.

You must subtract the negative area from the positive area to obtain net displacement.

Adding the absolute values would give total distance travelled instead.

These are different physical quantities:

• Distance tells you how much ground an object covers.
• Displacement tells you its overall change in position.

For meeting questions, you need displacement because you need the final position.

For a more detailed explanation of these graphical relationships, see Velocity-Time Graphs: Slope, Area and Motion Explained.

05Three details that can change your answer

Before moving into practice, there are three situations worth checking.

Position is not always the same as displacement

A position-time graph uses a shared spatial reference to show where objects are.

A displacement-time graph can instead show how far each object has moved from its own starting position.

Consider two runners:

• Runner A starts at \(x=0\ \mathrm{m}\).
• Runner B starts at \(x=10\ \mathrm{m}\).

Both move \(5\ \mathrm{m}\) to the right.

Their displacement values are both \(+5\ \mathrm{m}\), but their positions are:

\[ x_A=5\ \mathrm{m} \]

\[ x_B=15\ \mathrm{m} \]

Their displacement-time graphs have the same value, but the runners have not met.

You must add each object’s initial position before comparing their actual positions.

This is one reason it is important to understand the difference between position and displacement. See Displacement-Time vs Displacement-Position Graphs in HSC Physics.

The graphs must use the same reference frame

An intersection only establishes a meeting if both graphs use:

• The same time origin.
• The same position origin.
• The same positive direction.
• Compatible position units and time units.

If these conditions are not satisfied, you need to convert the data before comparing the graphs.

Real objects have dimensions

In many HSC problems, objects are treated as particles. This means their physical dimensions are ignored.

For particle models, meeting means their positions are equal.

Real objects, however, can touch without their centres occupying the same position.

For two carts moving towards each other, their front surfaces can touch while their centres remain separated.

This distinction becomes important in collision questions, where the objects’ velocities may change at contact.

06A reliable method for meeting questions

When you encounter a question involving two objects and graphical motion, use this sequence.

1. Check the axes. Identify whether the graph shows position, displacement, or velocity against time.

2. Check the reference frame. Make sure both objects use the same position origin and time origin.

3. If given position-time graphs, look for an intersection within the relevant time interval.

4. If given velocity-time graphs, calculate the signed area for each object and add it to its initial position.

5. Compare positions at the same time. Equal positions mean the tracked points meet.

6. Check the slopes. Different slopes mean different instantaneous velocities. Equal slopes are not required.

7. Consider physical contact. If the objects have significant dimensions or collide, check whether their motion changes before their position graphs would intersect.

The final step is particularly useful in harder HSC questions. A mathematical intersection predicted by constant-velocity equations is not necessarily physically possible if a collision occurs first.

07Questions and solutions

The following questions build from straightforward position-time calculations to more demanding combinations of motion graphs, signed areas, reference frames, and momentum.

Question 1

Two cyclists travel towards each other along a straight road.

At \(t=0\):

• Cyclist A is at \(x=4.0\ \mathrm{m}\) and travels right at \(3.0\ \mathrm{m\,s^{-1}}\).
• Cyclist B is at \(x=24.0\ \mathrm{m}\) and travels left at \(2.0\ \mathrm{m\,s^{-1}}\).

Take right as the positive direction.

Assuming constant velocities, calculate:

(a) The time at which they meet.

(b) Their position when they meet.

(c) The slope of each cyclist’s position-time graph.

Solution 1

The cyclists meet after \(4.0\ \mathrm{s}\), at \(x=16.0\ \mathrm{m}\). Their position-time graph slopes are \(+3.0\) and \(-2.0\ \mathrm{m\,s^{-1}}\), respectively.

Step 1: Write the position equations.

For cyclist A:

\[ x_A=4.0+3.0t \]

For cyclist B:

\[ x_B=24.0-2.0t \]

Step 2: Set the positions equal.

\[ 4.0+3.0t=24.0-2.0t \]

Rearranging:

\[ 5.0t=20.0 \]

Therefore:

\[ \boxed{t=4.0\ \mathrm{s}} \]

Step 3: Calculate their meeting position.

Substitute \(t=4.0\ \mathrm{s}\):

\[ x_A=4.0+3.0(4.0) \]

\[ \boxed{x=16.0\ \mathrm{m}} \]

Step 4: Identify the graph slopes.

The slope of a position-time graph is velocity.

Therefore:

\[ \boxed{m_A=+3.0\ \mathrm{m\,s^{-1}}} \]

\[ \boxed{m_B=-2.0\ \mathrm{m\,s^{-1}}} \]

The cyclists meet despite having different slopes because their positions become equal at the same instant.

Question 2

Two runners move along adjacent parallel lanes. Their positions are described by the following equations, with position in metres and time in seconds:

\[ x_A=1+t \]

\[ x_B=7-4t+t^2 \]

The equations are valid for \(0\leq t\leq4\).

(a) Determine every time at which the runners have the same longitudinal position.

(b) Calculate their positions at each meeting.

(c) Determine the instantaneous velocity of each runner at both meetings.

(d) Explain how runner B can meet runner A twice, even though runner B changes direction during the motion.

Solution 2

The runners have the same longitudinal position twice: at \(t=2.0\ \mathrm{s}\), where \(x=3.0\ \mathrm{m}\), and at \(t=3.0\ \mathrm{s}\), where \(x=4.0\ \mathrm{m}\).

Their velocities are different at both meetings.

Step 1: Set the position equations equal.

\[ 1+t=7-4t+t^2 \]

Rearrange:

\[ t^2-5t+6=0 \]

Factorise:

\[ (t-2)(t-3)=0 \]

Therefore:

\[ \boxed{t=2.0\ \mathrm{s}\text{ or }3.0\ \mathrm{s}} \]

Both times lie within the specified interval.

Step 2: Calculate the positions.

At \(t=2.0\ \mathrm{s}\):

\[ x_A=1+2=3.0\ \mathrm{m} \]

At \(t=3.0\ \mathrm{s}\):

\[ x_A=1+3=4.0\ \mathrm{m} \]

Because the positions are equal at each time, these are both meetings.

Step 3: Calculate the instantaneous velocities.

The velocity of runner A is constant:

\[ v_A=+1.0\ \mathrm{m\,s^{-1}} \]

The instantaneous gradient of runner B’s position-time curve is:

\[ v_B=-4+2t \]

At the first meeting:

\[ v_B=-4+2(2) \]

\[ \boxed{v_B=0\ \mathrm{m\,s^{-1}}} \]

At the second meeting:

\[ v_B=-4+2(3) \]

\[ \boxed{v_B=+2.0\ \mathrm{m\,s^{-1}}} \]

Step 4: Explain the two meetings.

At \(t=2.0\ \mathrm{s}\), runner B is at the turning point of their motion. Their velocity is zero at that instant.

Runner A continues moving right at \(1.0\ \mathrm{m\,s^{-1}}\).

After this instant, runner B reverses direction and accelerates towards the right.

At \(t=3.0\ \mathrm{s}\), runner B has caught runner A again.

The two meetings occur because the position-time curves intersect twice. The fact that runner B changes direction creates a curved graph, allowing multiple intersections.

Question 3

Two objects, A and B, move along a straight coordinate line. Assume their prescribed motions are unaffected by contact, so they can pass through one another.

Object A starts at \(x=0\ \mathrm{m}\) and moves at a constant velocity of \(+2.0\ \mathrm{m\,s^{-1}}\) for the first three seconds.

Object B starts at \(x=8.0\ \mathrm{m}\) and has the following velocity:

• From \(t=0\) to \(t=1.0\ \mathrm{s}\), its velocity is \(-2.0\ \mathrm{m\,s^{-1}}\).
• From \(t=1.0\) to \(t=3.0\ \mathrm{s}\), its velocity increases uniformly from \(-2.0\) to \(+2.0\ \mathrm{m\,s^{-1}}\).
• After \(t=3.0\ \mathrm{s}\), it continues at \(+2.0\ \mathrm{m\,s^{-1}}\).

(a) Determine whether the objects meet within the first three seconds.

(b) If they meet, calculate the meeting time and position.

(c) Determine their velocities at the meeting.

(d) Explain why using the total distance travelled by B instead of its displacement would produce an incorrect answer.

Solution 3

The objects first meet at \(t=3.0\ \mathrm{s}\), at \(x=6.0\ \mathrm{m}\). Both have a velocity of \(+2.0\ \mathrm{m\,s^{-1}}\) at that instant.

The important detail is that B’s negative and positive displacement areas cancel during the second interval.

Step 1: Calculate A’s position at \(t=3.0\ \mathrm{s}\).

Object A moves at a constant positive velocity.

\[ \Delta x_A=vt \]

\[ \Delta x_A=2.0(3.0) \]

\[ \Delta x_A=+6.0\ \mathrm{m} \]

Therefore:

\[ x_A=6.0\ \mathrm{m} \]

Step 2: Calculate B’s displacement during the first second.

The velocity is constant at \(-2.0\ \mathrm{m\,s^{-1}}\).

\[ \Delta x_{B1}=(-2.0)(1.0) \]

\[ \Delta x_{B1}=-2.0\ \mathrm{m} \]

Its position at \(t=1.0\ \mathrm{s}\) is:

\[ x_B=8.0-2.0=6.0\ \mathrm{m} \]

Step 3: Calculate B’s displacement from \(t=1\) to \(t=3\).

The velocity increases uniformly from \(-2.0\) to \(+2.0\ \mathrm{m\,s^{-1}}\).

There are two triangular areas:

• One below the time axis, representing negative displacement.
• One above the time axis, representing positive displacement.

The negative area is:

\[ A_-=-\frac12(1.0)(2.0) \]

\[ A_-=-1.0\ \mathrm{m} \]

The positive area is:

\[ A_+=+\frac12(1.0)(2.0) \]

\[ A_+=+1.0\ \mathrm{m} \]

Therefore:

\[ \Delta x_{B2}=-1.0+1.0 \]

\[ \boxed{\Delta x_{B2}=0\ \mathrm{m}} \]

Step 4: Calculate B’s final position.

Its total displacement is:

\[ \Delta x_B=-2.0+0 \]

\[ \Delta x_B=-2.0\ \mathrm{m} \]

Its final position is:

\[ x_B=8.0-2.0 \]

\[ x_B=6.0\ \mathrm{m} \]

Both objects are therefore at \(x=6.0\ \mathrm{m}\) at \(t=3.0\ \mathrm{s}\).

Step 5: Check that they have not met earlier.

During the first second, the separation decreases from \(8.0\ \mathrm{m}\) to \(4.0\ \mathrm{m}\), so they have not met.

For the second interval, let \(\tau=t-1\), where \(0\leq\tau\leq2\).

The positions are:

\[ x_A=2+2\tau \]

\[ x_B=6-2\tau+\tau^2 \]

Their separation is:

\[ x_B-x_A=4-4\tau+\tau^2 \]

Factorising:

\[ x_B-x_A=(2-\tau)^2 \]

This is positive for \(0\leq\tau<2\), and becomes zero only at \(\tau=2\).

Therefore, the first meeting occurs at \(t=3.0\ \mathrm{s}\).

Step 6: Determine their velocities at the meeting.

Object A:

\[ v_A=+2.0\ \mathrm{m\,s^{-1}} \]

Object B has reached \(+2.0\ \mathrm{m\,s^{-1}}\) by \(t=3.0\ \mathrm{s}\).

Therefore:

\[ \boxed{v_A=v_B=+2.0\ \mathrm{m\,s^{-1}}} \]

They meet with equal velocities and continue together at the same position if B maintains its specified velocity after three seconds.

Why the distance approach fails:

Object B travels:

• \(2.0\ \mathrm{m}\) to the left during the first second.
• \(1.0\ \mathrm{m}\) to the left during the next interval.
• \(1.0\ \mathrm{m}\) to the right during the next interval.

Its total distance is \(4.0\ \mathrm{m}\), but its displacement is only \(-2.0\ \mathrm{m}\).

Using distance instead of displacement would give the wrong final position.

This question also reveals something interesting: objects can meet even when their position-time graphs have equal slopes at the intersection. Different slopes are allowed, but they are not a requirement.

Question 4

Two video trackers record the motion of two runners separately along the same straight track.

Tracker A uses:

• A position origin at the western end of the track.
• A time origin at the beginning of the motion.

Its equation for runner A is:

\[ x_A=0.50+1.0t \]

Tracker B uses a different reference system:

• Its position origin is \(1.0\ \mathrm{m}\) east of tracker A’s position origin.
• Its clock starts \(0.50\ \mathrm{s}\) after tracker A’s clock.

Tracker B reports runner B’s position as:

\[ x'_B=3.0-2.0\tau \]

Here, \(x'_B\) is measured relative to tracker B’s position origin, and \(\tau\) is measured using tracker B’s clock.

Assume both position models remain valid throughout the relevant interval.

(a) Convert runner B’s position equation into tracker A’s coordinate system.

(b) Calculate the actual meeting time and position using tracker A’s reference system.

(c) A student overlays the two original equations without converting their coordinates or time origins. Explain why the apparent intersection is incorrect.

Solution 4

The actual meeting occurs at \(t=1.50\ \mathrm{s}\) and \(x=2.00\ \mathrm{m}\) in tracker A’s coordinate system.

The original equations cannot be compared directly because their position and time references differ.

Step 1: Convert the time variable.

Tracker B’s clock starts \(0.50\ \mathrm{s}\) after tracker A’s clock.

Therefore:

\[ \tau=t-0.50 \]

Step 2: Convert the position coordinate.

Tracker B’s position origin is \(1.0\ \mathrm{m}\) east of tracker A’s origin.

Therefore:

\[ x_B=x'_B+1.0 \]

Step 3: Substitute the coordinate and time conversions.

Start with:

\[ x'_B=3.0-2.0\tau \]

Substitute \(\tau=t-0.50\):

\[ x'_B=3.0-2.0(t-0.50) \]

Now convert to tracker A’s position coordinate:

\[ x_B=3.0-2.0(t-0.50)+1.0 \]

Simplifying:

\[ \boxed{x_B=5.0-2.0t} \]

Both equations now use the same reference system.

Step 4: Find the meeting time.

Runner A has:

\[ x_A=0.50+1.0t \]

Set the positions equal:

\[ 0.50+1.0t=5.0-2.0t \]

Rearranging:

\[ 3.0t=4.50 \]

Therefore:

\[ \boxed{t=1.50\ \mathrm{s}} \]

Step 5: Find the meeting position.

Substitute into runner A’s equation:

\[ x_A=0.50+1.0(1.50) \]

\[ \boxed{x=2.00\ \mathrm{m}} \]

Runner B’s clock reads:

\[ \tau=1.50-0.50 \]

\[ \tau=1.00\ \mathrm{s} \]

Both trackers therefore identify the same physical event once their measurements are converted correctly.

Step 6: Explain the student’s mistake.

If the student incorrectly treats \(x'_B=x_B\) and \(\tau=t\), they obtain:

\[ 0.50+t=3.0-2.0t \]

This gives:

\[ t=0.833\ \mathrm{s} \]

But this is not a valid meeting time because the two equations refer to different position and time origins.

The apparent intersection is a result of comparing incompatible coordinates.

The lesson is that a graph intersection has physical meaning only when the axes represent the same quantities using compatible reference systems.

Question 5

Two carts move towards each other along a straight, horizontal track.

They have the following properties:

The carts have sticky bumpers, so they remain attached after their first contact.

Assume negligible external horizontal forces and that the collision occurs over a negligible time interval.

(a) Calculate the time at which the carts first make contact.

(b) Calculate the position of each cart’s centre at contact.

(c) Calculate their common velocity immediately after the collision using conservation of momentum.

(d) Explain why the centre-position graphs do not intersect, even though the carts physically meet.

(e) A student extends both original straight-line centre-position graphs and predicts that the centres will meet at a later time. Explain precisely why this prediction is physically invalid.

Solution 5

The carts first make contact after \(1.33\ \mathrm{s}\). Their centres are then separated by \(0.50\ \mathrm{m}\), and they stick together while moving right at \(0.25\ \mathrm{m\,s^{-1}}\).

This is the important distinction: physical contact does not require the centres of extended objects to occupy the same position.

Step 1: Calculate the centre-to-centre separation required for contact.

Cart A has a length of \(0.40\ \mathrm{m}\), so its centre is \(0.20\ \mathrm{m}\) from its front surface.

Cart B has a length of \(0.60\ \mathrm{m}\), so its centre is \(0.30\ \mathrm{m}\) from its front surface.

When the carts touch:

\[ d_{\text{contact}}=0.20+0.30 \]

\[ \boxed{d_{\text{contact}}=0.50\ \mathrm{m}} \]

Step 2: Calculate the initial centre separation.

\[ d_0=3.00-0.50 \]

\[ d_0=2.50\ \mathrm{m} \]

The centres must therefore approach each other by:

\[ \Delta d=2.50-0.50 \]

\[ \Delta d=2.00\ \mathrm{m} \]

Step 3: Calculate their relative closing speed.

Cart A moves right at \(+1.0\ \mathrm{m\,s^{-1}}\).

Cart B moves left at \(-0.50\ \mathrm{m\,s^{-1}}\).

Their closing speed is:

\[ v_{\text{closing}}=1.0-(-0.50) \]

\[ \boxed{v_{\text{closing}}=1.50\ \mathrm{m\,s^{-1}}} \]

Step 4: Calculate the contact time.

Using distance divided by closing speed:

\[ t=\frac{2.00}{1.50} \]

\[ \boxed{t=1.33\ \mathrm{s}} \]

More precisely:

\[ t=\frac43\ \mathrm{s} \]

Step 5: Calculate the centre positions at contact.

For cart A:

\[ x_A=0.50+1.0\left(\frac43\right) \]

\[ \boxed{x_A=1.83\ \mathrm{m}} \]

For cart B:

\[ x_B=3.00-0.50\left(\frac43\right) \]

\[ \boxed{x_B=2.33\ \mathrm{m}} \]

Check their separation:

\[ 2.33-1.83=0.50\ \mathrm{m} \]

This matches the separation required for contact.

The facing surfaces occupy the same position when the carts touch, even though their centres do not.

Step 6: Calculate the common velocity after impact.

The carts stick together, so they have a common final velocity.

Apply conservation of momentum:

\[ m v_A+m v_B=(m+m)v_f \]

Substitute the initial velocities:

\[ m(1.0)+m(-0.50)=2mv_f \]

\[ 0.50m=2mv_f \]

Cancel \(m\):

\[ v_f=\frac{0.50}{2} \]

Therefore:

\[ \boxed{v_f=+0.25\ \mathrm{m\,s^{-1}}} \]

The attached carts continue moving to the right.

Step 7: Explain what happens to their centre-position graphs.

Before contact, their centre-position equations are:

\[ x_A=0.50+1.0t \]

\[ x_B=3.00-0.50t \]

At \(t=\frac43\ \mathrm{s}\), the carts collide.

After the collision, both centres have the same velocity:

\[ v_A=v_B=+0.25\ \mathrm{m\,s^{-1}} \]

Their positions therefore change according to:

\[ x_A=1.83+0.25\left(t-\frac43\right) \]

\[ x_B=2.33+0.25\left(t-\frac43\right) \]

Their centre positions remain separated by \(0.50\ \mathrm{m}\).

Their position-time graphs become parallel after contact. They never intersect.

Step 8: Identify the student’s mistake.

The student extends the original centre-position equations beyond the collision:

\[ 0.50+1.0t=3.00-0.50t \]

Solving gives:

\[ 1.50t=2.50 \]

\[ t=1.67\ \mathrm{s} \]

However, this predicted intersection occurs after the carts have already collided and stuck together at \(1.33\ \mathrm{s}\).

The original velocity equations are no longer valid after contact.

The student has mathematically extended two straight lines but ignored the physical event that changes their motion.

This is the crucial distinction:

• If you track the facing surfaces, their position-time graphs intersect when the carts touch.
• If you track their centres, their graphs do not intersect because the carts have finite lengths.
• After a collision, the gradients may change because the velocities change.

The objects have physically met even though their centre-position graphs never cross.

08Where to go next

The central rule is simple: a meeting is determined by position at the same time, not by the gradients of the position-time graphs.

Different gradients mean different velocities. They do not prevent an intersection.

When a question gives you velocity-time graphs, use signed areas to calculate displacement, then add the initial positions. When it involves actual collisions, check whether the objects’ dimensions or changing velocities affect the result.

These ideas connect directly to relative velocity, acceleration, and conservation of momentum, which are the natural next steps when analysing two interacting objects in HSC Physics.