Can a Velocity-Time Graph Cross Zero While an Object Keeps Moving?

A car travels to the right, slows down, reaches zero velocity, and then travels to the left. Its velocity-time graph crosses the horizontal axis. Has the car stopped? Yes, but only for an instant. It can be moving immediately before and immediately after that instant, even though its velocity is exactly zero at the crossing.

This is an important distinction in HSC Physics: **an object can be moving on both sides of a zero-velocity crossing, but it cannot have non-zero speed at the exact instant when its velocity is zero.** If the graph crosses from positive to negative velocity, the object momentarily stops and reverses direction.

The key is to interpret the sign of velocity, not just the shape of the graph.

## What does crossing zero actually mean?

Start with a simple sign convention:

- Positive velocity means motion to the right.
- Negative velocity means motion to the left.
- Zero velocity means the object's instantaneous speed is zero.

Imagine a car travelling along a straight road. It approaches a turnaround point, slows down, briefly stops, and reverses direction.

[IMAGE: A car travelling along a straight horizontal road, shown in three successive positions: moving right while slowing down, momentarily stationary at a turnaround point, and moving left after reversing direction. Include clear direction arrows and make the middle position visibly stationary.]

The car has not remained stationary for a noticeable period. It has simply reached an instant at which its velocity is zero.

This is what a continuous velocity-time graph crossing the time axis represents.

[IMAGE: An annotated velocity-time graph with time on the horizontal axis and signed velocity on the vertical axis. Show a straight line descending from positive velocity, crossing v = 0 at t = 4 s, and continuing into negative velocity. Label the regions as motion to the right, instantaneous rest and motion to the left. Mark the crossing point clearly.]

For example, consider the following illustrative data:

| Time (s) | Velocity (m/s) | Interpretation |
|---:|---:|---|
| 0 | +12 | Moving right |
| 1 | +9 | Moving right |
| 2 | +6 | Moving right |
| 3 | +3 | Moving right |
| 4 | 0 | Momentarily stationary |
| 5 | −3 | Moving left |
| 6 | −6 | Moving left |

The velocity changes continuously from positive to negative. At \(t = 4\text{ s}\), the car is momentarily at rest. Before that time, it moves right; afterwards, it moves left.

Notice that zero velocity does not mean zero acceleration. The car can still have a non-zero acceleration while its velocity is zero. In this example, it continues to accelerate towards the left throughout the turnaround.

## Two graphs that look similar but describe different motion

A useful way to avoid mistakes is to compare a graph that crosses zero with one that only touches zero.

### Case 1: The graph crosses the time axis

[IMAGE: A velocity-time graph with a straight line descending through the horizontal axis from positive to negative velocity. Annotate the positive-velocity region, the zero-velocity instant and the negative-velocity region. Include direction arrows showing a reversal.]

When velocity changes sign, the direction of motion changes.

- Above the time axis: positive velocity.
- At the time axis: zero instantaneous velocity.
- Below the time axis: negative velocity.

The object reverses direction at the crossing, provided the graph represents continuous velocity.

### Case 2: The graph touches the time axis and turns back

[IMAGE: A smooth U-shaped velocity-time graph that starts above the time axis, touches v = 0 at its minimum, and rises back into positive velocity without entering the negative region. Clearly distinguish touching from crossing.]

Here, the velocity reaches zero but remains positive on either side of the instant.

The object momentarily stops, but it does not reverse direction. It continues moving in the same direction afterwards.

This is the subtle distinction:

| Graph behaviour | What happens to velocity? | What happens to direction? |
|---|---|---|
| Crosses zero | Changes sign | Reverses |
| Touches zero and returns to the same side | Reaches zero but keeps the same sign | Does not reverse |
| Lies along zero for an interval | Remains zero | Object is stationary throughout that interval |

A graph crossing zero is therefore different from a graph merely reaching zero. Both can represent an instant of rest, but only the crossing represents a change in direction.

## The speed-time graph tells a different story

There is another useful way to check your interpretation.

Speed is the magnitude of velocity. It cannot be negative.

For one-dimensional motion,

\[
\text{speed}=|v|
\]

Suppose a car's velocity changes from \(+8\ \text{m s}^{-1}\) to \(-8\ \text{m s}^{-1}\). Its speed decreases from 8 to 0, then increases from 0 to 8.

The velocity graph crosses the axis, but the speed graph touches the axis and turns upwards.

[IMAGE: Side-by-side comparison of a velocity-time graph and a speed-time graph for the same car reversing direction. The velocity graph is a straight descending line from +8 m/s through zero to -8 m/s. The speed graph forms a V shape, decreasing from 8 m/s to zero and then increasing to 8 m/s. Use matching time axes and clearly label both vertical axes.]

This gives you an important distinction:

- A velocity-time graph can have positive and negative values.
- A speed-time graph cannot have negative values.
- If the object's full velocity is zero, its speed is also zero.
- If only one velocity component is zero, the object's total speed may still be positive.

Don't confuse a graph's slope with its height, either. A negative slope means negative acceleration, not necessarily negative velocity.

## How a motion sensor tests the interpretation

Consider a practical investigation.

A trolley rolls up a slightly inclined track, slows down, stops momentarily, and rolls back towards its starting point.

You want to determine exactly what happens to its velocity during the turnaround.

### The apparatus and the original measurement problem

Place an ultrasonic motion sensor at the lower end of the track, pointing uphill towards a reflector attached to the trolley. Connect the sensor to a data logger or computer.

The sensor sends out sound pulses and detects their reflections. From the time taken for the echoes to return, the system estimates the distance between the sensor and the trolley.

[IMAGE: Side-view experimental apparatus showing an ultrasonic motion sensor at the bottom of a gently inclined track, aimed towards a trolley with a reflector. Show the trolley travelling uphill away from the sensor and its subsequent return. Include the sensor's connection to a computer interface and label the measured distance as increasing away from the sensor.]

A stopwatch and several marked positions could give you average speeds between points. However, that method makes it difficult to identify the precise instant of the turnaround.

The motion sensor provides a better approach. It records position at short time intervals, allowing software to estimate velocity from the change in position.

Choose the positive direction to be away from the sensor, uphill.

This means:

- As the trolley moves uphill, its measured distance increases and its velocity is positive.
- As it slows down, the rate of increase in distance becomes smaller.
- At its furthest position, its instantaneous velocity is zero.
- As it rolls back, its distance from the sensor decreases and its velocity becomes negative.

### What should the measurements show?

Suppose an idealised investigation produces the following data. The trolley begins 0.50 m from the sensor and moves uphill with an initial velocity of \(+0.80\ \text{m s}^{-1}\). Its acceleration is constant at \(-0.40\ \text{m s}^{-2}\).

| Time (s) | Distance from sensor (m) | Velocity (m/s) |
|---:|---:|---:|
| 0 | 0.50 | +0.80 |
| 1 | 1.10 | +0.40 |
| 2 | 1.30 | 0.00 |
| 3 | 1.10 | −0.40 |

These are illustrative idealised readings, rather than measurements from an actual experiment.

Notice the pattern. The trolley reaches its greatest distance from the sensor at \(t=2\ \text{s}\). Its velocity is zero at that instant, and its velocity becomes negative afterwards.

[IMAGE: Two vertically aligned graphs for the illustrative trolley experiment. The upper distance-time graph rises from 0.50 m to a maximum of 1.30 m at 2 s, then falls to 1.10 m at 3 s. The lower velocity-time graph is a straight descending line from +0.80 m/s at 0 s, through zero at 2 s, to -0.40 m/s at 3 s. Align the time axes and clearly mark the turnaround.]

The observation supports the prediction: the trolley moves away, reaches an instantaneous turning point, and returns.

### What could make the result misleading?

A real motion sensor does not measure instantaneous velocity directly. It measures distance at discrete time intervals, and the software estimates velocity using those readings.

This creates two limitations:

1. **Finite sampling:** The actual turnaround can occur between recorded measurements.
2. **Measurement uncertainty:** Small fluctuations in measured distance can make the estimated velocity jump slightly above or below zero.

For this reason, don't assume that one displayed zero is automatically a perfect measurement of the turning instant. Look for the broader pattern: the position reaches a maximum and the estimated velocity changes sign.

The experiment also illustrates why direction matters. A sensor measuring distance alone gives positive distance readings throughout. The direction of motion comes from how that distance changes with time.

## The important exception: projectile motion

So far, we've considered motion along one straight line. In two-dimensional motion, the answer needs another layer.

A velocity vector can be split into horizontal and vertical components.

- \(v_x\) is the horizontal component of velocity.
- \(v_y\) is the vertical component of velocity.

The total speed is calculated using

\[
v=\sqrt{v_x^2+v_y^2}
\]

Now consider a basketball travelling through the air after being thrown towards a hoop.

At its highest point, the ball's vertical velocity is zero. However, it still has horizontal velocity.

[IMAGE: A basketball following a parabolic projectile trajectory towards a hoop. Show velocity-component arrows at three positions: on the way up, at the highest point, and on the way down. At the highest point, draw a horizontal velocity arrow and no vertical velocity arrow. Include a downward gravitational acceleration arrow at the highest point.]

At the highest point:

\[
v_y=0
\]

But:

\[
v_x\neq 0
\]

Therefore, the total velocity is not zero.

The ball continues moving horizontally while its vertical motion changes from upwards to downwards.

This is a genuine example of a velocity-time graph crossing zero while the object keeps moving with non-zero speed.

The crucial detail is that the graph represents only the vertical component of velocity.

### Worked example: Is a projectile stationary at its highest point?

A ball is launched at an angle with an initial horizontal velocity of \(12\ \text{m s}^{-1}\) and an initial vertical velocity of \(+8.0\ \text{m s}^{-1}\). Ignore air resistance and take gravitational acceleration as \(10\ \text{m s}^{-2}\) downwards.

Determine:

- The time taken to reach maximum height.
- The ball's total speed at maximum height.
- Whether the ball is stationary at that instant.

**Step 1**: Find the vertical velocity equation.

Taking upwards as positive, the initial vertical velocity is \(+8.0\ \text{m s}^{-1}\), and the vertical acceleration is \(-10\ \text{m s}^{-2}\).

Using

\[
v_y=u_y+a_yt
\]

gives

\[
v_y=8.0-10t
\]

**Step 2**: Find when the vertical velocity becomes zero.

At maximum height, the vertical component of velocity is zero.

\[
0=8.0-10t
\]

\[
t=0.80\ \text{s}
\]

The ball reaches maximum height after \(0.80\ \text{s}\).

**Step 3**: Determine the horizontal velocity.

With air resistance ignored, horizontal acceleration is zero. Therefore, the horizontal velocity remains constant:

\[
v_x=12\ \text{m s}^{-1}
\]

At maximum height, the velocity components are

\[
v_x=12\ \text{m s}^{-1}
\]

\[
v_y=0
\]

**Step 4**: Calculate the total speed.

\[
v=\sqrt{v_x^2+v_y^2}
\]

\[
v=\sqrt{12^2+0^2}
\]

\[
v=12\ \text{m s}^{-1}
\]

The ball is still travelling at \(12\ \text{m s}^{-1}\) at its highest point.

It has not stopped. Only its vertical component of velocity has reached zero.

Its acceleration is also still \(10\ \text{m s}^{-2}\) downwards. Zero vertical velocity does not mean zero vertical acceleration.

## What does the area around the crossing tell you?

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

This is particularly useful when an object reverses direction because the positive and negative regions contribute differently.

- Area above the time axis represents positive displacement.
- Area below the time axis represents negative displacement.
- Adding the signed areas gives net displacement.
- Adding the magnitudes of the areas gives total distance travelled.

[IMAGE: An annotated velocity-time graph showing a straight line descending from +0.80 m/s at 0 s through zero at 2 s to -0.40 m/s at 3 s. Shade the positive triangular region separately from the negative triangular region. Label the first area +0.80 m and the second area -0.20 m.]

### Worked example: How far does the trolley travel?

A trolley initially moves away from a motion sensor with a velocity of \(+0.80\ \text{m s}^{-1}\). Its velocity decreases uniformly to zero after \(2.0\ \text{s}\), then reaches \(-0.40\ \text{m s}^{-1}\) after a further \(1.0\ \text{s}\).

Calculate its total displacement and total distance travelled over the three seconds.

**Step 1**: Divide the graph at the zero-velocity crossing.

There are two regions:

- A positive triangular area from 0 to 2.0 s.
- A negative triangular area from 2.0 to 3.0 s.

**Step 2**: Calculate the positive displacement.

The first region is a triangle.

\[
A_1=\frac{1}{2}bh
\]

\[
A_1=\frac{1}{2}(2.0)(0.80)
\]

\[
A_1=+0.80\ \text{m}
\]

**Step 3**: Calculate the negative displacement.

The second region is also a triangle.

\[
A_2=-\frac{1}{2}(1.0)(0.40)
\]

\[
A_2=-0.20\ \text{m}
\]

The negative sign indicates displacement towards the sensor.

**Step 4**: Calculate the net displacement.

\[
\Delta x=A_1+A_2
\]

\[
\Delta x=0.80-0.20
\]

\[
\boxed{\Delta x=+0.60\ \text{m}}
\]

The trolley finishes 0.60 m further uphill than its starting position.

**Step 5**: Calculate the total distance.

Distance is the sum of the magnitudes of both areas.

\[
d=|A_1|+|A_2|
\]

\[
d=0.80+0.20
\]

\[
\boxed{d=1.00\ \text{m}}
\]

The trolley travels 1.00 m in total, but its net displacement is only 0.60 m because it reverses direction.

This is why you must keep the signs when calculating displacement, but remove them when calculating total distance.

If you need to revise the complete method for finding acceleration from slope and displacement from signed area, see [Velocity-Time Graphs: Slope, Area and Motion Explained](https://tutorgum.com/learn/resources/hsc-physics-velocity-time-graphs/).

For the related question of how an acceleration-time graph gives velocity change, see [Acceleration-Time Graphs: Area and Velocity Change](https://tutorgum.com/learn/resources/hsc-physics-acceleration-time-graphs/).

## A reliable way to interpret a zero crossing

When you encounter a velocity-time graph that reaches or crosses zero in an HSC question, use this sequence.

**1. Check what the vertical axis represents.**

Is it the full velocity along a straight line, or a component such as \(v_x\) or \(v_y\)?

**2. Identify the sign before and after zero.**

For one-dimensional motion:

- Positive to negative means a reversal of direction.
- Negative to positive also means a reversal.
- Positive to positive means the object touches zero without reversing.
- Negative to negative means the object touches zero without reversing.

**3. Distinguish velocity from speed.**

Zero full velocity means zero speed. Zero velocity in just one direction does not necessarily mean zero total speed.

**4. Check the signed areas.**

Use signed areas to calculate displacement. Add the magnitudes of the areas to calculate total distance.

**5. Check acceleration separately.**

The graph's height represents velocity. Its slope represents acceleration. An object can have zero velocity and non-zero acceleration at the same instant.

## Questions and solutions

Try each question before reading its solution. The questions progress from interpreting a simple crossing to combining signed areas and two-dimensional motion.

### Question 1

A car travels along a straight road. Its velocity-time graph is a straight line from \(+8.0\ \text{m s}^{-1}\) at \(t=0\) to \(-4.0\ \text{m s}^{-1}\) at \(t=6.0\ \text{s}\).

[IMAGE: A velocity-time graph with a straight descending line joining (0 s, +8 m/s) and (6 s, -4 m/s). Show the horizontal zero-velocity axis and label the time and velocity axes. Do not shade or calculate the areas.]

Determine:

a) The time at which the car changes direction.

b) The car's direction of motion before and after this time.

c) Its displacement over the six seconds.

d) Its total distance travelled.

### Solution 1

The car changes direction at \(t=4.0\ \text{s}\), has a displacement of \(+12\ \text{m}\), and travels a total distance of \(20\ \text{m}\).

**Part a: Find the zero-velocity time.**

The graph has a constant gradient, so acceleration is constant.

\[
a=\frac{v_f-v_i}{\Delta t}
\]

\[
a=\frac{-4.0-8.0}{6.0}
\]

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

Using \(v=u+at\):

\[
0=8.0-2.0t
\]

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

**Part b: Determine the direction.**

Before \(4.0\ \text{s}\), velocity is positive, so the car travels in the positive direction.

After \(4.0\ \text{s}\), velocity is negative, so it travels in the opposite direction.

The car momentarily stops and reverses direction at \(4.0\ \text{s}\).

**Part c: Calculate displacement.**

The positive triangular area is

\[
A_+=\frac{1}{2}(4.0)(8.0)
\]

\[
A_+=+16\ \text{m}
\]

The negative triangular area is

\[
A_-=-\frac{1}{2}(2.0)(4.0)
\]

\[
A_-=-4.0\ \text{m}
\]

Therefore,

\[
\Delta x=16-4
\]

\[
\boxed{\Delta x=+12\ \text{m}}
\]

**Part d: Calculate total distance.**

Add the magnitudes of both areas.

\[
d=16+4
\]

\[
\boxed{d=20\ \text{m}}
\]

The car travels 20 m but finishes only 12 m from its starting position in the positive direction.

### Question 2

A trolley has a velocity-time graph consisting of two straight sections:

- From \(t=0\) to \(t=3.0\ \text{s}\), its velocity decreases uniformly from \(+6.0\ \text{m s}^{-1}\) to zero.
- From \(t=3.0\) to \(t=5.0\ \text{s}\), its velocity increases uniformly from zero to \(+4.0\ \text{m s}^{-1}\).

[IMAGE: A piecewise linear velocity-time graph starting at +6 m/s at 0 s, descending to touch the horizontal axis at 3 s, then rising to +4 m/s at 5 s. Keep both sections above or on the axis. Mark the turning point without adding a direction arrow that gives away the answer.]

A student claims that the trolley reverses direction at \(t=3.0\ \text{s}\) because its velocity-time graph has a negative slope before that time.

Assess this claim. Calculate:

a) The acceleration during each section.

b) The trolley's direction of motion before and after \(t=3.0\ \text{s}\).

c) Its total displacement over five seconds.

d) Its speed at \(t=4.0\ \text{s}\).

### Solution 2

The student is incorrect. The trolley momentarily stops at \(3.0\ \text{s}\), but it does not reverse direction. Its velocity remains positive on both sides of the zero.

**Part a: Calculate acceleration.**

For the first section:

\[
a_1=\frac{0-6.0}{3.0}
\]

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

For the second section:

\[
a_2=\frac{4.0-0}{5.0-3.0}
\]

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

The acceleration changes direction at \(3.0\ \text{s}\).

**Part b: Determine whether direction changes.**

Before \(3.0\ \text{s}\), velocity is positive.

After \(3.0\ \text{s}\), velocity is also positive.

Therefore, the trolley moves in the positive direction on both sides of the zero-velocity point.

The negative slope before \(3.0\ \text{s}\) indicates negative acceleration. It does not indicate negative velocity.

The trolley slows down, stops instantaneously, and then speeds up in its original direction.

**Part c: Calculate displacement.**

Both regions are above the time axis, so both areas are positive.

First triangular area:

\[
A_1=\frac{1}{2}(3.0)(6.0)
\]

\[
A_1=9.0\ \text{m}
\]

Second triangular area:

\[
A_2=\frac{1}{2}(2.0)(4.0)
\]

\[
A_2=4.0\ \text{m}
\]

Total displacement:

\[
\Delta x=9.0+4.0
\]

\[
\boxed{\Delta x=+13.0\ \text{m}}
\]

**Part d: Find the speed at \(4.0\ \text{s}\).**

During the second section, acceleration is \(+2.0\ \text{m s}^{-2}\).

The trolley starts that section from rest.

\[
v=u+at
\]

Here, the elapsed time since the start of the second section is \(1.0\ \text{s}\).

\[
v=0+(2.0)(1.0)
\]

\[
\boxed{v=+2.0\ \text{m s}^{-1}}
\]

Its speed is \(2.0\ \text{m s}^{-1}\).

The important point is that a negative acceleration does not automatically mean negative velocity or motion in the negative direction.

### Question 3

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

- It starts at \(+6.0\ \text{m s}^{-1}\) at \(t=0\).
- Its velocity decreases uniformly to zero at \(t=2.0\ \text{s}\).
- It then continues decreasing uniformly to \(-3.0\ \text{m s}^{-1}\) at \(t=6.0\ \text{s}\).

[IMAGE: A piecewise linear velocity-time graph joining (0 s, +6 m/s), (2 s, 0 m/s), and (6 s, -3 m/s). Clearly show the positive and negative triangular regions, but do not shade them or label their calculated areas.]

A student argues:

"The positive and negative areas cancel, so the trolley must have stopped at \(t=6.0\ \text{s}\)."

Determine:

a) The net displacement over the six seconds.

b) The total distance travelled.

c) The trolley's final velocity.

d) Whether the student's conclusion is correct.

### Solution 3

The net displacement is zero, but the trolley is not stationary at \(t=6.0\ \text{s}\). It has returned to its starting position while travelling in the negative direction at \(3.0\ \text{m s}^{-1}\).

**Part a: Calculate net displacement.**

The first region is a positive triangle:

\[
A_1=\frac{1}{2}(2.0)(6.0)
\]

\[
A_1=+6.0\ \text{m}
\]

The second region is a negative triangle:

\[
A_2=-\frac{1}{2}(4.0)(3.0)
\]

\[
A_2=-6.0\ \text{m}
\]

Therefore,

\[
\Delta x=6.0-6.0
\]

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

**Part b: Calculate total distance.**

Distance uses the magnitudes of both areas.

\[
d=|6.0|+|-6.0|
\]

\[
\boxed{d=12.0\ \text{m}}
\]

The trolley travels 12.0 m in total.

**Part c: Determine final velocity.**

The question gives the velocity at \(t=6.0\ \text{s}\).

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

The negative sign means the trolley is travelling in the negative direction.

**Part d: Assess the student's conclusion.**

The student's reasoning confuses zero displacement with zero velocity.

Zero net displacement means the trolley has returned to its starting position. It says nothing about whether the trolley is stationary at that moment.

In this case, the trolley reaches zero velocity at \(2.0\ \text{s}\), reverses direction, and eventually returns to its original position at \(6.0\ \text{s}\).

It is still moving when it arrives.

This is an important distinction: position, displacement, velocity, and speed describe different aspects of motion.

### Question 4

A ball is launched from ground level with the following initial velocity components:

- Horizontal velocity: \(+15\ \text{m s}^{-1}\).
- Vertical velocity: \(+20\ \text{m s}^{-1}\).

Ignore air resistance and take gravitational acceleration as \(10\ \text{m s}^{-2}\) downwards.

[IMAGE: A projectile launched diagonally upwards from ground level, showing its parabolic trajectory. Include horizontal and vertical velocity-component arrows at launch and at the highest point. At the highest point, show a horizontal arrow labelled 15 m/s and a downward acceleration arrow labelled 10 m/s^2. Do not show a vertical velocity arrow at the apex.]

Calculate:

a) The time taken to reach maximum height.

b) The maximum vertical height above the launch point.

c) The total speed at maximum height.

d) The acceleration at maximum height.

Explain whether the ball is stationary at the highest point.

### Solution 4

The ball reaches maximum height after \(2.0\ \text{s}\), rises \(20\ \text{m}\), and still has a horizontal speed of \(15\ \text{m s}^{-1}\). It is not stationary.

**Part a: Find the time to maximum height.**

Take upwards as positive.

The initial vertical velocity is

\[
u_y=+20\ \text{m s}^{-1}
\]

The vertical acceleration is

\[
a_y=-10\ \text{m s}^{-2}
\]

At maximum height, \(v_y=0\).

Using

\[
v_y=u_y+a_yt
\]

\[
0=20-10t
\]

\[
\boxed{t=2.0\ \text{s}}
\]

**Part b: Calculate maximum height.**

Use

\[
\Delta y=u_yt+\frac{1}{2}a_yt^2
\]

Substitute:

\[
\Delta y=(20)(2.0)+\frac{1}{2}(-10)(2.0)^2
\]

\[
\Delta y=40-20
\]

\[
\boxed{\Delta y=20\ \text{m}}
\]

The ball rises 20 m above its launch point.

**Part c: Calculate total speed at maximum height.**

The horizontal velocity remains constant because there is no horizontal acceleration.

\[
v_x=15\ \text{m s}^{-1}
\]

At maximum height:

\[
v_y=0
\]

The total speed is

\[
v=\sqrt{v_x^2+v_y^2}
\]

\[
v=\sqrt{15^2+0^2}
\]

\[
\boxed{v=15\ \text{m s}^{-1}}
\]

The ball continues moving horizontally.

**Part d: Determine acceleration.**

Gravity acts throughout the flight.

Therefore,

\[
\boxed{a_y=-10\ \text{m s}^{-2}}
\]

The acceleration is still directed downwards at maximum height.

The ball has zero vertical velocity, but non-zero horizontal velocity and non-zero vertical acceleration.

### Question 5

Two balls, A and B, are launched simultaneously from the same point.

Both have the same initial vertical velocity of \(+20\ \text{m s}^{-1}\). However:

- Ball A is thrown vertically upwards, with zero horizontal velocity.
- Ball B is launched diagonally upwards, with a horizontal velocity of \(+15\ \text{m s}^{-1}\).

Ignore air resistance and take gravitational acceleration as \(10\ \text{m s}^{-2}\) downwards.

[IMAGE: Two side-by-side projectile trajectories starting at the same origin. Ball A follows a vertical path upwards and back down. Ball B follows a parabolic path with constant horizontal progression. Show that both have identical vertical motion, marking their common apex time at 2 s and return time at 4 s. At the apex, show no velocity arrow for A and a horizontal velocity arrow for B. Include a small inset of their identical vertical velocity-time graphs, descending from +20 m/s at 0 s through zero at 2 s to -20 m/s at 4 s.]

A student examines the vertical velocity-time graphs and makes two claims:

1. Both balls must stop completely at \(t=2.0\ \text{s}\), because both graphs reach zero.
2. Both balls must return to their original launch positions at \(t=4.0\ \text{s}\), because the positive and negative areas under both vertical velocity-time graphs cancel.

Evaluate both claims.

Your answer must include:

a) The vertical velocity equation for each ball.

b) The total velocity and speed of each ball at \(t=2.0\ \text{s}\).

c) The vertical and horizontal displacements of each ball at \(t=4.0\ \text{s}\).

d) An explanation of why identical vertical velocity-time graphs do not necessarily describe identical overall motion.

### Solution 5

The student's first claim is only correct for ball A. Ball B continues moving horizontally at \(15\ \text{m s}^{-1}\) when its vertical velocity reaches zero.

The second claim is also incorrect. Both balls return to their original vertical height, but only ball A returns to its original horizontal position.

This problem exposes the limitation of interpreting a single velocity component as the object's entire motion.

**Part a: Find the vertical velocity equation.**

Both balls have the same initial vertical velocity:

\[
u_y=+20\ \text{m s}^{-1}
\]

Both experience the same vertical acceleration:

\[
a_y=-10\ \text{m s}^{-2}
\]

Therefore, for both balls:

\[
v_y=u_y+a_yt
\]

\[
\boxed{v_y=20-10t}
\]

Their vertical velocity-time graphs are identical.

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

\[
v_y=20-10(2.0)=0
\]

Both balls reach maximum height at the same time.

They also return to their original vertical height when

\[
0=20-10t
\]

giving the second solution

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

**Part b: Compare their total velocities at the highest point.**

For ball A, the horizontal velocity is zero throughout its flight.

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

\[
v_{x,A}=0
\]

\[
v_{y,A}=0
\]

Therefore,

\[
v_A=\sqrt{0^2+0^2}
\]

\[
\boxed{v_A=0\ \text{m s}^{-1}}
\]

Ball A is momentarily stationary at the highest point before reversing its vertical direction.

For ball B, horizontal velocity remains constant:

\[
v_{x,B}=+15\ \text{m s}^{-1}
\]

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

\[
v_{y,B}=0
\]

Therefore,

\[
v_B=\sqrt{15^2+0^2}
\]

\[
\boxed{v_B=15\ \text{m s}^{-1}}
\]

Ball B is not stationary. It continues moving horizontally.

The two balls have identical vertical velocity at every instant, but different total velocities.

**Part c: Calculate their displacements at \(t=4.0\ \text{s}\).**

First, consider vertical displacement.

For either ball:

\[
\Delta y=u_yt+\frac{1}{2}a_yt^2
\]

\[
\Delta y=(20)(4.0)+\frac{1}{2}(-10)(4.0)^2
\]

\[
\Delta y=80-80
\]

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

Both balls return to their original vertical height.

Now calculate horizontal displacement.

For ball A:

\[
\Delta x_A=v_{x,A}t
\]

\[
\Delta x_A=(0)(4.0)
\]

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

Ball A returns to its original launch point.

For ball B:

\[
\Delta x_B=v_{x,B}t
\]

\[
\Delta x_B=(15)(4.0)
\]

\[
\boxed{\Delta x_B=+60\ \text{m}}
\]

Ball B lands 60 m horizontally from its launch point.

**Part d: Explain the limitation of the graphs.**

The student's mistake is assuming that a vertical velocity-time graph describes the object's entire velocity.

It does not.

The area under each vertical velocity-time graph gives vertical displacement only.

The positive triangular area from 0 to 2.0 s is

\[
A_+=\frac{1}{2}(2.0)(20)=+20\ \text{m}
\]

The negative triangular area from 2.0 to 4.0 s is

\[
A_-=-\frac{1}{2}(2.0)(20)=-20\ \text{m}
\]

The areas cancel:

\[
\Delta y=+20-20=0\ \text{m}
\]

But this only establishes that the balls return to their original vertical height.

It does not establish that their horizontal displacements are zero.

The same reasoning applies to the zero-velocity crossing at the apex. For ball A, both components are zero, so its total speed is zero. For ball B, only the vertical component is zero.

The exact same vertical velocity-time graph can therefore describe an object that momentarily stops and another object that never stops at all.

This is why you must always check whether a graph represents the full velocity or just one component.

---

The next useful step is connecting velocity components to projectile trajectories. Once you can distinguish horizontal and vertical motion, you can use their separate velocity-time graphs to predict displacement, maximum height, and landing position without confusing a momentary zero component with the object coming to a complete stop.