Sharp Corners on Motion Graphs: What Changes and Why

A trolley is moving to the right at \(+8.0\text{ m s}^{-1}\). Its velocity-time graph reaches a sharp corner and then slopes downwards.

Does the trolley’s velocity suddenly drop at the corner?

No. Its velocity remains \(+8.0\text{ m s}^{-1}\) at that instant. What changes abruptly is the gradient of the graph, which means its acceleration changes.

There is an important catch, though. The meaning of a sharp corner depends on which quantity is plotted on the vertical axis.

• On a velocity-time graph, a sharp corner means acceleration changes abruptly.
• On a displacement-time graph, a sharp corner means velocity changes abruptly.
• At the exact mathematical corner, the gradient is not uniquely defined.

Let’s work through why these statements are different, what they mean physically, and how to handle the signed areas when a velocity-time graph changes direction.

01First, identify which graph has the corner

A sharp corner is a point where a graph changes gradient suddenly without jumping to a different value.

The distinction between the plotted quantity and its gradient is crucial.

There is another feature worth distinguishing: a vertical jump.

A vertical jump on a velocity-time graph means velocity itself changes instantaneously. That is different from a corner in the graph.

The easiest way to remember this is:

A corner changes the gradient. A vertical jump changes the plotted quantity.

Let’s concentrate on the velocity-time graph first, because this is where the distinction between acceleration, force, and impulse becomes particularly useful.

02What happens at a corner on a velocity-time graph?

Imagine a trolley moving along a straight track.

Initially, a constant net force pushes it to the right. The trolley accelerates, so its velocity increases.

Now suppose the net force suddenly changes direction and points to the left.

What happens to the trolley’s velocity?

It does not suddenly jump to a smaller value. The trolley is still moving to the right at the instant the force changes.

Instead, it begins accelerating to the left. Its rightward velocity starts decreasing.

This is exactly what a sharp corner on a velocity-time graph represents.

The gradient tells us acceleration

The gradient of a velocity-time graph is acceleration:

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

where:

• \(a\) is acceleration in metres per second squared (\(\text{m s}^{-2}\)).
• \(\Delta v\) is the change in velocity in metres per second (\(\text{m s}^{-1}\)).
• \(\Delta t\) is the time interval in seconds (s).

Before the corner, the graph has one gradient. After the corner, it has another.

We can describe these as:

\[ a_{\text{before}}\ne a_{\text{after}} \]

The velocity, however, is continuous:

\[ v_{\text{before}}=v_{\text{after}} \]

at the corner.

This means the object does not suddenly gain or lose velocity. It simply begins changing velocity at a different rate.

What happens to the net force?

Newton’s second law gives:

\[ F_{\text{net}}=ma \]

where \(F_{\text{net}}\) is net force in newtons (N), \(m\) is mass in kilograms (kg), and \(a\) is acceleration in \(\text{m s}^{-2}\).

For an object with constant mass, a sudden change in acceleration corresponds to a sudden change in net force.

However, the force does not have to become infinite.

For example, the net force might change from \(+6\text{ N}\) to \(-4\text{ N}\). Both values are finite. The force simply changes from one value to another.

This is an idealised model of a force switching abruptly.

Is acceleration defined at the exact corner?

Not if the left-hand and right-hand gradients are different.

At the corner, there are two possible gradients:

• The gradient immediately before the corner.
• The gradient immediately after the corner.

Neither gives a unique instantaneous gradient at the exact point.

Therefore, acceleration is mathematically undefined at that instant in the idealised model.

In an HSC response, it is better to state the acceleration before and after the corner rather than assigning one acceleration to the corner itself.

Importantly, this does not mean acceleration is infinite. The acceleration values on either side can both be finite.

03How would we observe this in an experiment?

Consider a proposed experiment using a trolley on a low-friction track.

A string connects the trolley to a small hanging mass over a pulley. The hanging mass pulls the trolley horizontally.

An ultrasonic motion sensor sits at one end of the track, facing the trolley. It records the trolley’s position at regular time intervals.

At a chosen time, the string is released, removing the horizontal pulling force.

This is a proposed measurement design, rather than a report of measured experimental results.

What should the velocity-time graph look like?

Before the string is released, the hanging mass produces a net horizontal force on the trolley.

Assuming this force is approximately constant, the trolley has constant positive acceleration. Its velocity-time graph rises linearly.

When the string is released, the horizontal net force becomes approximately zero, assuming friction is negligible.

The acceleration becomes approximately zero, so the trolley continues at constant velocity.

The idealised velocity-time graph therefore has a sharp corner:

• Before release: positive gradient.
• At release: velocity remains continuous.
• After release: zero gradient.

Why might the experimental graph look less sharp?

The motion sensor cannot measure continuously at every possible instant.

Suppose it records data at 10 readings per second. The time between readings is:

\[ \Delta t=0.10\text{ s} \]

Now imagine the string is released \(0.03\text{ s}\) after one velocity reading.

Suppose the trolley has an acceleration of \(+1.0\text{ m s}^{-2}\) before release, followed by zero acceleration.

During the first \(0.03\text{ s}\), its velocity changes by:

\[ \Delta v=a\Delta t \]

\[ \Delta v=1.0(0.03) \]

\[ \Delta v=0.030\text{ m s}^{-1} \]

If we estimate acceleration over the entire \(0.10\text{ s}\) sampling interval, we obtain:

\[ a_{\text{average}}=\frac{0.030}{0.10} \]

\[ a_{\text{average}}=0.30\text{ m s}^{-2} \]

But the actual acceleration in this simplified model was \(+1.0\text{ m s}^{-2}\) before release and zero afterwards.

The calculated \(0.30\text{ m s}^{-2}\) is an average across an interval that contains two different motion conditions.

This is a limitation of the original measurement design.

A better design would use a higher sampling rate and record the release signal alongside the motion data. This would help locate the transition more accurately.

Even then, an experiment cannot establish a perfectly instantaneous change. Real forces take some time to change, and measurements have finite resolution.

The sharp corner is a useful mathematical idealisation of a transition that happens over a very short time.

04Signed areas still work when the graph has a corner

Now for the other important part of the problem.

A corner changes the gradient of a velocity-time graph, but it does not change the rule for calculating displacement.

The signed area under a velocity-time graph represents displacement.

For a straight-line section, the area can be calculated using a trapezium:

\[ \Delta x=\frac{v_1+v_2}{2}\Delta t \]

where:

• \(\Delta x\) is displacement in metres (m).
• \(v_1\) and \(v_2\) are the velocities at the beginning and end of the interval.
• \(\Delta t\) is the duration of the interval in seconds.

The units work:

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

The important detail is that areas are signed.

For total distance, use the magnitudes of the areas instead.

A sharp corner does not contribute a special area of its own. It simply tells you where one straight-line section ends and another begins.

05The contrasting case: a corner on a displacement-time graph

Now let’s change the graph.

Suppose a displacement-time graph has two straight-line sections meeting at a sharp corner.

Remember that the gradient of a displacement-time graph represents velocity:

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

Imagine an object moving from \(x=0\text{ m}\) at \(t=0\text{ s}\) to \(x=+4\text{ m}\) at \(t=2\text{ s}\). It then moves to \(x=0\text{ m}\) at \(t=4\text{ s}\).

The velocity before the corner is:

\[ v_1=\frac{4-0}{2-0} \]

\[ v_1=+2\text{ m s}^{-1} \]

After the corner:

\[ v_2=\frac{0-4}{4-2} \]

\[ v_2=-2\text{ m s}^{-1} \]

The object has changed from moving right at \(2\text{ m s}^{-1}\) to moving left at \(2\text{ m s}^{-1}\).

But there is a problem.

The change happens instantaneously in the idealised graph.

Why does this require an impulse?

Velocity has changed by:

\[ \Delta v=v_2-v_1 \]

\[ \Delta v=-2-(+2) \]

\[ \Delta v=-4\text{ m s}^{-1} \]

Suppose the object has a mass of \(0.50\text{ kg}\).

The impulse is:

\[ J=\Delta p=m\Delta v \]

where \(J\) is impulse in newton seconds (\(\text{N s}\)), and \(\Delta p\) is the change in momentum.

Substituting:

\[ J=(0.50)(-4) \]

\[ \boxed{J=-2.0\text{ N s}} \]

The negative sign indicates that the impulse acts in the negative direction.

A finite net force cannot produce this velocity change in zero time. The idealised model requires an impulsive force: a force acting over a very short time, with a finite total impulse.

This is a useful model for an extremely brief collision.

In a real collision, the force acts over a small but nonzero time. The velocity changes rapidly but continuously, so the displacement-time graph would have a rounded transition rather than a mathematically perfect corner.

What happens to the signed area?

The corresponding velocity-time graph has:

• A constant velocity of \(+2\text{ m s}^{-1}\) for the first \(2\text{ s}\).
• A constant velocity of \(-2\text{ m s}^{-1}\) for the next \(2\text{ s}\).

The first area is:

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

The second area is:

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

Therefore:

\[ \Delta x=4-4=0\text{ m} \]

This matches the displacement-time graph: the object finishes where it started.

The vertical step in the velocity-time graph has zero width in time, so it contributes no additional displacement.

This is an important distinction:

• A corner on a velocity-time graph changes acceleration.
• A corner on a displacement-time graph changes velocity.
• A vertical jump on a velocity-time graph represents an instantaneous velocity change.

06A quick decision process for HSC questions

When you encounter a sharp corner, work through these steps.

1. Identify the vertical axis.

Is the graph showing displacement, velocity, or acceleration?

2. Compare the gradients on either side.

Calculate the gradient before and after the corner separately.

3. Check whether the plotted quantity is continuous.

A corner normally means the plotted value remains continuous. A vertical jump means it does not.

4. Translate the gradient into physics.

• Displacement-time gradient gives velocity.
• Velocity-time gradient gives acceleration.
• For constant mass, acceleration determines net force.

5. If it is a velocity-time graph, calculate signed areas.

Split the graph at corners and zero crossings. Add positive and negative areas for displacement. Add their magnitudes for distance.

The most common mistake is to assume that a change in gradient means the plotted quantity itself must have changed instantaneously. It doesn’t.

07Questions and solutions

Question 1

A \(1.5\text{ kg}\) trolley moves along a straight track. Its velocity is constant at \(+4.0\text{ m s}^{-1}\) from \(t=0\) to \(t=2.0\text{ s}\). It then decreases uniformly to \(+1.0\text{ m s}^{-1}\) at \(t=5.0\text{ s}\).

Take right as positive.

Calculate the acceleration and net force before and after the corner. State the direction in which the trolley is moving at the corner.

Solution 1

The acceleration is \(0\text{ m s}^{-2}\) before the corner and \(-1.0\text{ m s}^{-2}\) afterwards. The corresponding net forces are \(0\text{ N}\) and \(-1.5\text{ N}\). The trolley is still moving right at the corner.

Before the corner:

\[ a_1=\frac{4.0-4.0}{2.0} \]

\[ a_1=0\text{ m s}^{-2} \]

Therefore:

\[ F_1=ma_1 \]

\[ F_1=(1.5)(0) \]

\[ \boxed{F_1=0\text{ N}} \]

After the corner:

\[ a_2=\frac{1.0-4.0}{5.0-2.0} \]

\[ a_2=\frac{-3.0}{3.0} \]

\[ a_2=-1.0\text{ m s}^{-2} \]

The net force is:

\[ F_2=(1.5)(-1.0) \]

\[ \boxed{F_2=-1.5\text{ N}} \]

At \(t=2.0\text{ s}\), the velocity is still \(+4.0\text{ m s}^{-1}\).

The trolley therefore continues moving right, but begins slowing down.

The exact acceleration at the corner is undefined because the left-hand and right-hand gradients are different.

Question 2

A \(2.0\text{ kg}\) cart moves along a straight horizontal track. Its velocity-time graph consists of two straight-line sections joining the following points:

Take right as positive.

Determine:

a) The acceleration during each interval.

b) The net force during each interval.

c) The total displacement from \(0\) to \(5.0\text{ s}\).

d) The total impulse acting on the cart over the full interval.

e) Whether the sharp corner at \(t=2.0\text{ s}\) represents an impulse.

Solution 2

The accelerations are \(+2.0\text{ m s}^{-2}\) and \(-2.0\text{ m s}^{-2}\). The net forces are \(+4.0\text{ N}\) and \(-4.0\text{ N}\), and the total displacement is \(+22\text{ m}\). The total impulse is \(-4.0\text{ N s}\). The corner itself does not represent an impulse.

Part a: Acceleration

For the first interval:

\[ a_1=\frac{7.0-3.0}{2.0} \]

\[ \boxed{a_1=+2.0\text{ m s}^{-2}} \]

For the second interval:

\[ a_2=\frac{1.0-7.0}{5.0-2.0} \]

\[ a_2=\frac{-6.0}{3.0} \]

\[ \boxed{a_2=-2.0\text{ m s}^{-2}} \]

Part b: Net force

Using \(F_{\text{net}}=ma\):

\[ F_1=(2.0)(+2.0) \]

\[ \boxed{F_1=+4.0\text{ N}} \]

And:

\[ F_2=(2.0)(-2.0) \]

\[ \boxed{F_2=-4.0\text{ N}} \]

The net force changes direction at the corner.

Part c: Displacement

From \(0\) to \(2.0\text{ s}\):

\[ \Delta x_1=\frac12(3.0+7.0)(2.0) \]

\[ \Delta x_1=+10\text{ m} \]

From \(2.0\) to \(5.0\text{ s}\):

\[ \Delta x_2=\frac12(7.0+1.0)(3.0) \]

\[ \Delta x_2=+12\text{ m} \]

Therefore:

\[ \Delta x=10+12 \]

\[ \boxed{\Delta x=+22\text{ m}} \]

Part d: Total impulse

Impulse equals change in momentum:

\[ J=m(v-u) \]

Substituting the initial and final velocities:

\[ J=2.0(1.0-3.0) \]

\[ \boxed{J=-4.0\text{ N s}} \]

We can also verify this using the force-time intervals:

\[ J=(4.0)(2.0)+(-4.0)(3.0) \]

\[ J=8.0-12 \]

\[ J=-4.0\text{ N s} \]

Part e: Does the corner produce an impulse?

No. The velocity is \(+7.0\text{ m s}^{-1}\) immediately before and after the corner.

The net force changes from \(+4.0\text{ N}\) to \(-4.0\text{ N}\), but there is no instantaneous change in momentum.

The tempting mistake is to treat every sudden force change as an impulse. An impulse requires a change in momentum over the relevant time interval. A bounded force switching between two values does not produce a finite impulse at a single instant.

Question 3

A puck moves along a straight track. Its displacement-time graph consists of two straight-line sections:

• At \(t=0\text{ s}\), its displacement is \(0\text{ m}\).
• At \(t=2.0\text{ s}\), its displacement is \(+6.0\text{ m}\).
• At \(t=5.0\text{ s}\), its displacement is \(-3.0\text{ m}\).

The puck has a mass of \(0.50\text{ kg}\).

a) Calculate the velocity before and after the corner.

b) Determine the change in velocity.

c) Calculate the impulse required by this idealised motion.

d) Explain whether the graph is physically possible if the puck experiences only finite forces.

e) Verify the final displacement using signed areas on the corresponding velocity-time graph.

Solution 3

The idealised graph requires an instantaneous velocity change from \(+3.0\text{ m s}^{-1}\) to \(-3.0\text{ m s}^{-1}\), corresponding to an impulse of \(-3.0\text{ N s}\). It is not physically possible under a finite-force-only model.

Part a: Velocity before and after the corner

The velocity is the gradient of the displacement-time graph.

Before the corner:

\[ v_1=\frac{6.0-0}{2.0-0} \]

\[ \boxed{v_1=+3.0\text{ m s}^{-1}} \]

After the corner:

\[ v_2=\frac{-3.0-6.0}{5.0-2.0} \]

\[ v_2=\frac{-9.0}{3.0} \]

\[ \boxed{v_2=-3.0\text{ m s}^{-1}} \]

Part b: Change in velocity

\[ \Delta v=v_2-v_1 \]

\[ \Delta v=-3.0-(+3.0) \]

\[ \boxed{\Delta v=-6.0\text{ m s}^{-1}} \]

Part c: Impulse

Using:

\[ J=m\Delta v \]

\[ J=(0.50)(-6.0) \]

\[ \boxed{J=-3.0\text{ N s}} \]

The impulse acts in the negative direction.

Part d: Physical interpretation

The displacement remains continuous at the corner, but the velocity changes instantaneously.

For a finite mass, this requires an impulsive force in the idealised model. A finite force acting for zero time cannot produce a finite change in momentum.

A real collision would take a small but nonzero amount of time, producing a rapid, continuous change in velocity.

Part e: Verify using signed areas

The corresponding velocity-time graph has a constant velocity of \(+3.0\text{ m s}^{-1}\) from \(0\) to \(2.0\text{ s}\), followed by \(-3.0\text{ m s}^{-1}\) from \(2.0\) to \(5.0\text{ s}\).

The first area is:

\[ A_1=(3.0)(2.0)=+6.0\text{ m} \]

The second area is:

\[ A_2=(-3.0)(3.0)=-9.0\text{ m} \]

Therefore:

\[ \Delta x=6.0-9.0 \]

\[ \boxed{\Delta x=-3.0\text{ m}} \]

This agrees with the displacement-time graph.

The vertical step in the velocity-time graph contributes no additional area because it has zero width in time.

Question 4

A \(2.0\text{ kg}\) cart moves along a straight track. Its velocity-time graph consists of two straight-line sections joining these points:

Take right as positive.

Determine:

a) The acceleration and net force during each interval.

b) The times at which the cart reverses direction.

c) The total displacement.

d) The total distance travelled.

e) Whether the corner at \(t=2.0\text{ s}\) represents a reversal or an impulse.

Solution 4

The accelerations are \(+2.0\text{ m s}^{-2}\) and \(-2.0\text{ m s}^{-2}\), giving net forces of \(+4.0\text{ N}\) and \(-4.0\text{ N}\). The cart reverses at \(t=1.0\text{ s}\) and \(t=3.0\text{ s}\). Its displacement is zero, but it travels a total distance of \(4.0\text{ m}\).

Part a: Acceleration and net force

For the first interval:

\[ a_1=\frac{2.0-(-2.0)}{2.0} \]

\[ \boxed{a_1=+2.0\text{ m s}^{-2}} \]

The net force is:

\[ F_1=(2.0)(2.0) \]

\[ \boxed{F_1=+4.0\text{ N}} \]

For the second interval:

\[ a_2=\frac{-2.0-2.0}{2.0} \]

\[ \boxed{a_2=-2.0\text{ m s}^{-2}} \]

The net force is:

\[ F_2=(2.0)(-2.0) \]

\[ \boxed{F_2=-4.0\text{ N}} \]

Part b: Times of reversal

The cart reverses whenever its velocity changes sign.

During the first interval, velocity increases from \(-2.0\) to \(+2.0\text{ m s}^{-1}\).

The gradient is \(+2.0\text{ m s}^{-2}\), so it takes \(1.0\text{ s}\) to increase from \(-2.0\) to zero.

Therefore, the first reversal occurs at:

\[ \boxed{t=1.0\text{ s}} \]

During the second interval, velocity decreases from \(+2.0\) to \(-2.0\text{ m s}^{-1}\).

It takes another \(1.0\text{ s}\) to reach zero.

Therefore, the second reversal occurs at:

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

Part c: Total displacement

From \(0\) to \(2.0\text{ s}\), the negative triangular area and positive triangular area cancel.

Equivalently:

\[ \Delta x_1=\frac12(-2.0+2.0)(2.0)=0 \]

From \(2.0\) to \(4.0\text{ s}\):

\[ \Delta x_2=\frac12(2.0-2.0)(2.0)=0 \]

Therefore:

\[ \boxed{\Delta x_{\text{total}}=0\text{ m}} \]

The cart finishes where it started.

Part d: Total distance

Distance requires adding the magnitudes of all four triangular areas.

Each triangle has a base of \(1.0\text{ s}\) and a height of \(2.0\text{ m s}^{-1}\).

\[ A=\frac12(1.0)(2.0) \]

\[ A=1.0\text{ m} \]

There are four such regions:

\[ d=1.0+1.0+1.0+1.0 \]

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

Part e: Interpretation of the corner

The corner occurs at \(t=2.0\text{ s}\), when the velocity is \(+2.0\text{ m s}^{-1}\).

The cart is still moving right at that instant.

The acceleration and net force change direction, but velocity remains continuous. Therefore, the corner does not represent either a reversal or an impulse.

The reversals happen at the two zero crossings, not at the corner.

Question 5

This final question tests a subtle limitation of interpreting motion graphs from measurements.

A \(1.0\text{ kg}\) cart moves along a straight track. Its measured positions are:

Its velocity is also measured as \(+4.0\text{ m s}^{-1}\) at \(t=0\), \(t=2.0\text{ s}\), and \(t=4.0\text{ s}\).

Two possible velocity-time models are proposed.

Model A: The cart travels at a constant velocity of \(+4.0\text{ m s}^{-1}\) throughout the four seconds.

Model B: The velocity changes linearly between the following points:

Straight lines connect consecutive points.

A student argues:

Both models have the same measured positions and the same measured velocities at the specified times. Therefore, the cart must have travelled at constant velocity with zero net force throughout.

Assess the student’s conclusion.

Your response must:

a) Determine whether both models satisfy the measured data.

b) Calculate the acceleration and net force on each interval in Model B.

c) Calculate the cart’s displacement at \(t=1.0\text{ s}\) in Model B.

d) Explain whether the sharp corners in Model B produce impulses.

e) Identify one additional measurement that would distinguish the two models.

Solution 5

The student’s conclusion is incorrect. Both models satisfy all the stated measurements, but Model B has nonzero acceleration and nonzero net force over several intervals. The measurements do not uniquely determine the motion between recorded times.

This is the important conceptual point: knowing the displacement over an interval and the velocities at selected instants does not tell you the complete velocity-time graph.

Part a: Check whether both models satisfy the measurements

For Model A, velocity is constant:

\[ v=+4.0\text{ m s}^{-1} \]

Its displacement over two seconds is:

\[ \Delta x=(4.0)(2.0)=8.0\text{ m} \]

Over four seconds:

\[ \Delta x=(4.0)(4.0)=16.0\text{ m} \]

It therefore satisfies all the measurements.

For Model B, calculate the signed area from \(0\) to \(2.0\text{ s}\).

From \(0\) to \(0.5\text{ s}\):

\[ A_1=\frac12(4+6)(0.5)=2.5\text{ m} \]

From \(0.5\) to \(1.5\text{ s}\):

\[ A_2=\frac12(6+2)(1.0)=4.0\text{ m} \]

From \(1.5\) to \(2.0\text{ s}\):

\[ A_3=\frac12(2+4)(0.5)=1.5\text{ m} \]

Adding:

\[ \Delta x=2.5+4.0+1.5 \]

\[ \Delta x=8.0\text{ m} \]

The same pattern repeats from \(2.0\) to \(4.0\text{ s}\), giving another \(8.0\text{ m}\).

Therefore:

\[ \boxed{\Delta x_{0\rightarrow4}=16.0\text{ m}} \]

Model B also has velocity \(+4.0\text{ m s}^{-1}\) at \(t=0\), \(2.0\), and \(4.0\text{ s}\).

Both models satisfy every stated measurement.

Part b: Acceleration and net force in Model B

Calculate the gradient of each straight-line section.

From \(0\) to \(0.5\text{ s}\):

\[ a_1=\frac{6-4}{0.5}=+4.0\text{ m s}^{-2} \]

From \(0.5\) to \(1.5\text{ s}\):

\[ a_2=\frac{2-6}{1.0}=-4.0\text{ m s}^{-2} \]

From \(1.5\) to \(2.0\text{ s}\):

\[ a_3=\frac{4-2}{0.5}=+4.0\text{ m s}^{-2} \]

The same three gradients repeat from \(2.0\) to \(4.0\text{ s}\).

The acceleration sequence is therefore:

\[ +4.0,\ -4.0,\ +4.0,\ +4.0,\ -4.0,\ +4.0 \]

in \(\text{m s}^{-2}\).

Since the cart has a mass of \(1.0\text{ kg}\), Newton’s second law gives:

\[ F_{\text{net}}=ma \]

The corresponding net forces are:

\[ \boxed{+4.0,\ -4.0,\ +4.0,\ +4.0,\ -4.0,\ +4.0\text{ N}} \]

Model B therefore involves changing net forces, despite having the same measured endpoint velocities as Model A.

Part c: Find the displacement at \(t=1.0\text{ s}\)

For Model B, calculate the area from \(0\) to \(1.0\text{ s}\).

From \(0\) to \(0.5\text{ s}\):

\[ A_1=\frac12(4+6)(0.5) \]

\[ A_1=2.5\text{ m} \]

From \(0.5\) to \(1.0\text{ s}\), the velocity decreases linearly from \(6\) to \(4\text{ m s}^{-1}\).

\[ A_2=\frac12(6+4)(0.5) \]

\[ A_2=2.5\text{ m} \]

Therefore:

\[ x(1.0)=2.5+2.5 \]

\[ \boxed{x(1.0)=5.0\text{ m}} \]

Model A would give:

\[ x(1.0)=(4.0)(1.0) \]

\[ \boxed{x(1.0)=4.0\text{ m}} \]

The models predict different positions at \(t=1.0\text{ s}\).

Part d: Do the corners produce impulses?

No. Model B has continuous velocity at every corner.

At each corner, the acceleration changes between finite values of \(+4.0\) and \(-4.0\text{ m s}^{-2}\).

The corresponding net force changes between \(+4.0\) and \(-4.0\text{ N}\), but there is no instantaneous change in velocity.

The force is not infinite, and there is no impulsive change in momentum at those corners.

There is another interesting feature here.

Over the first two seconds, the cart finishes with the same velocity with which it started:

\[ \Delta v=4.0-4.0=0 \]

Therefore, the total impulse over that interval is zero.

However, this does not mean the net force was zero throughout the interval.

The positive and negative impulses from the different force intervals cancel.

A zero total impulse does not require zero force at every instant.

Part e: What additional measurement would distinguish the models?

A position measurement at \(t=1.0\text{ s}\) would be sufficient.

Model A predicts \(4.0\text{ m}\), while Model B predicts \(5.0\text{ m}\).

The models are therefore distinguishable with one additional position reading.

The tempting mistake is to assume that equal average velocities and equal endpoint velocities prove constant velocity.

They do not.

The signed area under the velocity-time graph constrains displacement, but many different graph shapes can produce the same area. Similarly, knowing the initial and final velocities constrains the total change in momentum, but not the complete force history.

This is why you must distinguish what a graph actually establishes from what you are assuming about the motion between measurements.

08Where to go next

Once you can interpret sharp corners, the next useful step is translating between force-time, acceleration-time, velocity-time, and displacement-time graphs.

Remember the chain:

\[ F_{\text{net}}\rightarrow a\rightarrow v\rightarrow x \]

Each graph reveals a different part of the motion:

• Force determines acceleration.
• Acceleration determines how velocity changes.
• Velocity determines how displacement changes.
• Signed areas under a velocity-time graph give displacement.

For a broader explanation of the connections between force and motion, read Force, Motion and Graph Representations for HSC Physics.

If you want to strengthen your understanding of the contrasting case, Displacement-Time Graphs: How to Read Motion in HSC Physics covers how gradients reveal velocity and what changes in those gradients tell you about motion.