Can an Object Have Zero Velocity and Nonzero Acceleration?
A ball at the highest point of its flight has zero instantaneous velocity but still accelerates downwards. Learn why this happens, how to interpret velocity and acceleration signs, and how to solve related HSC Physics problems.
Yes. An object can have zero velocity and nonzero acceleration. A ball thrown vertically upwards provides the clearest example: at its highest point, its instantaneous velocity is zero, but its acceleration is still approximately \(9.8\,\mathrm{m\,s^{-2}}\) downwards, assuming air resistance is negligible.
The ball stops moving upwards for an instant. Gravity does not stop acting. Understanding why requires separating two ideas: how an object’s position is changing and how its velocity is changing.
01Why zero velocity does not mean zero acceleration
Imagine throwing a ball straight upwards.
As the ball rises, it slows down. Eventually, it reaches a highest point, where it momentarily stops moving upwards. It then starts falling.
Consider what happens immediately before and after this turning point:
- Before reaching the highest point, the ball has an upward velocity.
- At the highest point, its instantaneous velocity is zero.
- Immediately afterwards, it has a downward velocity.
Its velocity changes continuously from positive to negative (if upwards is defined as positive).
However, the acceleration remains directed downwards throughout the flight.
This is because acceleration describes the rate of change of velocity, not the object’s current speed.
At the highest point, the ball’s velocity is zero, but gravity is still accelerating it downwards.
Instantaneously stationary versus remaining at rest
There is an important difference between an object having zero velocity at one instant and remaining stationary over a period of time.
| Situation | Velocity | Acceleration |
|---|---|---|
| Ball at the highest point of its flight | Zero at one instant | Downwards |
| Ball held stationary on a table | Zero throughout the interval | Zero |
| Ball travelling upwards under gravity | Upwards | Downwards |
| Ball falling under gravity | Downwards | Downwards |
An object that remains stationary throughout a time interval has zero acceleration during that interval.
But an object that is stationary for just one instant can have nonzero acceleration.
The ball at the highest point is an example of the second situation.
02A video-tracking experiment: What happens at the highest point?
How could we distinguish between an object that has genuinely stopped accelerating and one that has simply reached zero velocity?
We can investigate this using a ball thrown vertically upwards.
Setting up the experiment
Place a vertical metre ruler behind the ball’s path. Position a camera a short distance away, facing the ruler directly.
The camera should record the ball moving upwards and downwards. The ruler must be in the same plane as the ball’s motion to minimise parallax error.
Record the video at a known frame rate, then track the centre of the ball at equal time intervals.
A stopwatch and visual observation might tell us approximately when the ball reaches its highest point. However, they cannot establish whether acceleration is zero at that instant.
Video tracking gives us something much more useful: a sequence of positions from which we can estimate changes in velocity.
What should we expect to observe?
Consider two possible models.
Model 1: Acceleration becomes zero at the highest point.
If the ball enters a zero-acceleration state and remains in it, its velocity stays constant. Since its velocity is already zero, the ball should remain at the same height.
Model 2: Gravity continues acting at the highest point.
The ball’s velocity continues changing downwards. It reaches zero momentarily, then becomes negative as the ball starts falling.
The second model predicts a curved position-time relationship, with the ball moving downwards after reaching its highest point.
Illustrative video-tracking data
Suppose the highest point occurs at \(t=0\), and we record the ball’s vertical position relative to that point.
The following values are illustrative idealised data, not measurements from an actual experiment.
| Time relative to highest point (s) | Height relative to highest point (m) |
|---|---|
| -0.20 | -0.196 |
| -0.10 | -0.049 |
| 0.00 | 0.000 |
| +0.10 | -0.049 |
| +0.20 | -0.196 |
Notice two things.
First, the ball reaches its highest position at \(t=0\).
Second, its position continues changing after that instant. It has not remained stationary.
The equal heights at equal times before and after the highest point are also consistent with uniformly accelerated motion.
Estimating the acceleration
We can use the position measurements to estimate velocity on either side of the highest point.
Between \(t=-0.20\,\mathrm{s}\) and \(t=0\), the average velocity is:
\[ v_{-0.10}\approx \frac{0-(-0.196)}{0.20} \]
\[ v_{-0.10}\approx +0.98\,\mathrm{m\,s^{-1}} \]
Similarly, between \(t=0\) and \(t=+0.20\,\mathrm{s}\):
\[ v_{+0.10}\approx \frac{-0.196-0}{0.20} \]
\[ v_{+0.10}\approx -0.98\,\mathrm{m\,s^{-1}} \]
These average velocities estimate the instantaneous velocities at the midpoints of their respective intervals.
Now calculate the acceleration:
\[ a=\frac{v_f-v_i}{\Delta t} \]
\[ a=\frac{-0.98-(+0.98)}{0.20} \]
\[ \boxed{a=-9.8\,\mathrm{m\,s^{-2}}} \]
The negative sign indicates that acceleration is directed downwards.
The central estimate of velocity at the highest point is zero, because the positions on either side are equal.
We have therefore identified exactly what we were looking for: zero velocity at the highest point, alongside nonzero downward acceleration.
In a real experiment, finite frame intervals, measurement uncertainty, and air resistance affect the estimates. Fitting a uniform-acceleration model to several video frames gives a more reliable result than relying on one pair of measurements.
03Understanding the signs using uniform acceleration
We can now describe the ball’s motion mathematically.
For vertical motion, choose upwards as the positive direction.
This means:
- Upwards velocity is positive.
- Downwards velocity is negative.
- Gravitational acceleration is negative.
Near the Earth’s surface, we take the magnitude of gravitational acceleration as \(g=9.8\,\mathrm{m\,s^{-2}}\). Therefore:
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
The negative sign does not mean gravity is getting weaker. It simply tells us its direction relative to our chosen positive axis.
The three uniform-acceleration equations
For constant acceleration, we can use:
\[ v=u+at \]
\[ \Delta y=ut+\frac12at^2 \]
\[ v^2=u^2+2a\Delta y \]
Here:
- \(u\) is initial velocity, in \(\mathrm{m\,s^{-1}}\).
- \(v\) is final velocity, in \(\mathrm{m\,s^{-1}}\).
- \(a\) is constant acceleration, in \(\mathrm{m\,s^{-2}}\).
- \(t\) is elapsed time, in seconds.
- \(\Delta y\) is vertical displacement, in metres.
For the ball thrown upwards, the first equation is particularly useful:
\[ v=u-9.8t \]
At the highest point, \(v=0\).
Therefore:
\[ 0=u-9.8t \]
Rearranging:
\[ \boxed{t=\frac{u}{9.8}} \]
This gives the time taken to reach the highest point.
Notice that we have set velocity equal to zero, not acceleration.
The acceleration remains \(-9.8\,\mathrm{m\,s^{-2}}\).
What does the velocity-time graph show?
A velocity-time graph makes the relationship particularly clear.
For constant acceleration, the gradient of a velocity-time graph equals acceleration.
For a ball thrown upwards, the graph is a straight line with a negative gradient.
It crosses the horizontal axis when velocity reaches zero. However, the line does not become horizontal.
That nonzero gradient tells us that acceleration is still acting.
This is an important distinction:
- The vertical position of the graph represents velocity.
- The gradient represents acceleration.
- Crossing the horizontal axis means velocity is zero.
- A nonzero gradient at that crossing means acceleration is nonzero.
For more practice interpreting how acceleration changes velocity over a time interval, see Acceleration-Time Graphs: Area and Velocity Change.
When does velocity equal zero? Understanding roots
The equation \(v=u+at\) also helps us understand why the timing matters.
Setting \(v=0\):
\[ 0=u+at \]
Therefore:
\[ \boxed{t=-\frac{u}{a}} \]
This is the mathematical root of the velocity equation.
For a ball thrown upwards, \(u\) is positive and \(a\) is negative. The resulting time is positive, so the ball reaches zero velocity after launch.
But consider what happens if the initial velocity and acceleration have the same sign.
The calculated time is negative. This means the zero-velocity point occurred before the chosen starting time, not during the future motion being considered.
There is another important distinction: a zero of the displacement equation is not necessarily a zero of the velocity equation.
For a ball returning to its launch height:
\[ 0=ut+\frac12at^2 \]
Factorising:
\[ 0=t\left(u+\frac12at\right) \]
This gives two roots:
\[ t=0 \]
and
\[ t=-\frac{2u}{a} \]
The first root represents the launch time. The second represents the time when the ball returns to its original height.
Neither equation says that the ball must have zero velocity at those two times.
For a ball launched upwards and returning to its launch height, the velocity-zero root occurs halfway between those two position roots.
This is why you should always check what physical quantity an equation represents before interpreting its roots.
04Worked example 1: Finding the velocity-zero point
A ball is thrown vertically upwards from ground level with an initial velocity of \(14.0\,\mathrm{m\,s^{-1}}\). Neglect air resistance.
Calculate:
a) The time taken to reach the highest point.
b) The maximum height above the launch position.
c) The acceleration at the highest point.
Step 1
Take upwards as positive.
\[ u=+14.0\,\mathrm{m\,s^{-1}} \]
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
At the highest point:
\[ v=0 \]
Step 2
Use:
\[ v=u+at \]
Substitute:
\[ 0=14.0-9.8t \]
Rearrange:
\[ t=\frac{14.0}{9.8} \]
\[ \boxed{t=1.43\,\mathrm{s}} \]
The ball reaches its highest point approximately \(1.43\) seconds after launch.
Step 3
Use:
\[ v^2=u^2+2a\Delta y \]
Substitute:
\[ 0=14.0^2+2(-9.8)\Delta y \]
Rearrange:
\[ 19.6\Delta y=196 \]
\[ \boxed{\Delta y=10.0\,\mathrm{m}} \]
The ball rises \(10.0\) metres above its launch position.
Step 4
The ball’s acceleration is constant throughout its flight:
\[ \boxed{a=-9.8\,\mathrm{m\,s^{-2}}} \]
The ball has zero velocity at the highest point, but it still accelerates downwards.
This is the central idea: zero velocity does not require zero acceleration.
05Worked example 2: A ball thrown upwards from a balcony
A ball is thrown vertically upwards from a balcony \(18.0\) metres above the ground with an initial velocity of \(12.0\,\mathrm{m\,s^{-1}}\).
Neglect air resistance and take upwards as positive.
Calculate:
a) The time taken to reach the highest point.
b) The maximum height above the ground.
c) The time taken to reach the ground.
d) The velocity immediately before impact.
Step 1
The initial height is:
\[ y_0=18.0\,\mathrm{m} \]
The initial velocity is:
\[ u=+12.0\,\mathrm{m\,s^{-1}} \]
The acceleration is:
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
Step 2
At the highest point, \(v=0\).
Using:
\[ v=u+at \]
\[ 0=12.0-9.8t \]
\[ t=\frac{12.0}{9.8} \]
\[ \boxed{t=1.22\,\mathrm{s}} \]
Step 3
First, find the vertical displacement from the balcony.
Using:
\[ v^2=u^2+2a\Delta y \]
\[ 0=12.0^2+2(-9.8)\Delta y \]
\[ 19.6\Delta y=144 \]
\[ \Delta y=7.35\,\mathrm{m} \]
The balcony is already \(18.0\) metres above the ground.
Therefore:
\[ y_{\max}=18.0+7.35 \]
\[ \boxed{y_{\max}=25.3\,\mathrm{m}} \]
Step 4
This time, we need the displacement equation:
\[ y=y_0+ut+\frac12at^2 \]
At ground level, \(y=0\).
Substitute:
\[ 0=18.0+12.0t-4.90t^2 \]
Rearrange:
\[ 4.90t^2-12.0t-18.0=0 \]
Using the quadratic formula:
\[ t=\frac{12.0\pm\sqrt{496.8}}{9.80} \]
Therefore:
\[ t=3.50\,\mathrm{s} \]
or
\[ t=-1.05\,\mathrm{s} \]
We choose the positive root:
\[ \boxed{t=3.50\,\mathrm{s}} \]
Why discard the negative root?
The negative root describes a mathematical position before the chosen launch time. It is not a future impact during the motion described in the question.
Step 5
Return to:
\[ v=u+at \]
Substitute the impact time:
\[ v=12.0-9.8(3.50) \]
\[ \boxed{v=-22.3\,\mathrm{m\,s^{-1}}} \]
The negative sign indicates downward motion.
The ball hits the ground at approximately \(22.3\,\mathrm{m\,s^{-1}}\) downwards.
Notice that the ball had zero velocity at \(1.22\) seconds, but its acceleration remained \(-9.8\,\mathrm{m\,s^{-2}}\). It then continued downwards and eventually reached the ground.
06Questions and solutions
The following questions move from straightforward applications towards more demanding situations involving position, velocity, acceleration, and reference frames.
Question 1
A ball is thrown vertically upwards from level ground with an initial velocity of \(9.8\,\mathrm{m\,s^{-1}}\).
Take upwards as positive and neglect air resistance.
a) Calculate the time at which the ball’s velocity becomes zero.
b) State the acceleration at this instant.
c) Explain why the ball does not remain at rest.
Solution 1
The ball reaches zero velocity after \(1.00\) second, but its acceleration remains \(9.8\,\mathrm{m\,s^{-2}}\) downwards.
Step 1: Identify the known values.
\[ u=+9.8\,\mathrm{m\,s^{-1}} \]
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
At the highest point:
\[ v=0 \]
Step 2: Calculate the time.
Using:
\[ v=u+at \]
\[ 0=9.8-9.8t \]
\[ \boxed{t=1.00\,\mathrm{s}} \]
Step 3: State the acceleration.
The acceleration is constant throughout the flight:
\[ \boxed{a=-9.8\,\mathrm{m\,s^{-2}}} \]
Step 4: Explain why the ball continues moving.
At the highest point, gravity continues accelerating the ball downwards.
Its velocity therefore changes from zero to a negative value immediately after reaching the highest point.
The ball begins falling.
Question 2
A trolley moves along a straight horizontal track.
At \(t=0\), it has a velocity of \(+6.0\,\mathrm{m\,s^{-1}}\). It experiences a constant acceleration of \(-2.0\,\mathrm{m\,s^{-2}}\) for at least four seconds.
Take right as positive.
a) Calculate the time at which the trolley momentarily stops.
b) Calculate its displacement from its starting position at this instant.
c) Calculate its velocity after \(4.0\) seconds.
d) Determine its position relative to its starting point after \(4.0\) seconds. Explain how it can be moving left while still being to the right of its starting position.
Solution 2
The trolley momentarily stops after \(3.0\) seconds, when it is \(9.0\) metres to the right of its starting position. After \(4.0\) seconds, it is still \(8.0\) metres to the right of its starting position, but is moving left at \(2.0\,\mathrm{m\,s^{-1}}\).
Step 1: Calculate when its velocity becomes zero.
Use:
\[ v=u+at \]
At the stopping instant:
\[ 0=6.0-2.0t \]
Therefore:
\[ \boxed{t=3.0\,\mathrm{s}} \]
Step 2: Calculate the displacement at this instant.
Use:
\[ \Delta x=ut+\frac12at^2 \]
Substitute:
\[ \Delta x=6.0(3.0)+\frac12(-2.0)(3.0)^2 \]
\[ \Delta x=18.0-9.0 \]
\[ \boxed{\Delta x=+9.0\,\mathrm{m}} \]
The trolley is 9.0 metres to the right of its starting point.
Step 3: Calculate the velocity after four seconds.
\[ v=u+at \]
\[ v=6.0-2.0(4.0) \]
\[ \boxed{v=-2.0\,\mathrm{m\,s^{-1}}} \]
The negative sign means the trolley is now travelling left.
Step 4: Calculate its displacement after four seconds.
\[ \Delta x=ut+\frac12at^2 \]
\[ \Delta x=6.0(4.0)+\frac12(-2.0)(4.0)^2 \]
\[ \Delta x=24.0-16.0 \]
\[ \boxed{\Delta x=+8.0\,\mathrm{m}} \]
The trolley is still 8.0 metres to the right of its starting position.
This is an important distinction:
- Velocity tells us the direction in which the trolley is currently moving.
- Displacement tells us where it is relative to its starting position.
The trolley has reversed direction, but it has not yet travelled back past its starting point.
At \(t=3.0\) seconds, its velocity is zero while its acceleration is still \(-2.0\,\mathrm{m\,s^{-2}}\).
Question 3
Consider two separate trials involving identical balls.
In Trial A, a ball is thrown vertically upwards from point P with an initial velocity of \(+14.7\,\mathrm{m\,s^{-1}}\).
In Trial B, another ball is released from rest at the exact position where the first ball reaches its highest point. Both trials use the same gravitational acceleration, and air resistance is negligible.
a) Calculate the time taken for the ball in Trial A to reach its highest point.
b) Determine its maximum height above P.
c) Calculate the velocity of each ball \(0.40\) seconds after the ball in Trial B is released.
d) Calculate the displacement of each ball from their common starting position during this \(0.40\)-second interval.
e) Explain whether the balls have different subsequent motions because one was thrown upwards and the other was released from rest.
Solution 3
Both balls have exactly the same velocity and position \(0.40\) seconds after the second ball is released. Their different histories do not produce different subsequent motion when their positions, velocities, and accelerations are identical.
Step 1: Find when the first ball reaches its highest point.
For Trial A:
\[ u=14.7\,\mathrm{m\,s^{-1}} \]
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
At the highest point:
\[ v=0 \]
Using:
\[ v=u+at \]
\[ 0=14.7-9.8t \]
\[ \boxed{t=1.50\,\mathrm{s}} \]
Step 2: Find the maximum height above P.
Use:
\[ v^2=u^2+2a\Delta y \]
\[ 0=14.7^2-19.6\Delta y \]
\[ \Delta y=\frac{216.09}{19.6} \]
\[ \boxed{\Delta y=11.0\,\mathrm{m}} \]
The first ball reaches a height of approximately 11.0 metres above P.
Step 3: Compare the balls’ velocities after \(0.40\) seconds.
At the instant Trial B begins:
- Ball A is at the highest point.
- Ball A has zero velocity.
- Ball B has just been released from rest, so its velocity is also zero.
Both balls now have:
\[ u=0 \]
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
After \(0.40\) seconds:
\[ v=u+at \]
\[ v=0-9.8(0.40) \]
\[ \boxed{v=-3.92\,\mathrm{m\,s^{-1}}} \]
Both balls have the same velocity.
Step 4: Calculate their displacement.
Use:
\[ \Delta y=ut+\frac12at^2 \]
\[ \Delta y=0+\frac12(-9.8)(0.40)^2 \]
\[ \boxed{\Delta y=-0.784\,\mathrm{m}} \]
Both balls move 0.784 metres downwards from their common starting position.
Step 5: Explain their subsequent motion.
It might seem that the first ball should behave differently because it has just finished moving upwards, whereas the second ball has only just started falling.
However, this reasoning focuses on their histories rather than their current physical conditions.
At the instant they are compared, both balls have:
- The same position.
- The same velocity.
- The same acceleration.
Under the same uniform-acceleration conditions, their subsequent motions must therefore be identical.
This demonstrates that an object can have zero velocity and nonzero acceleration, regardless of whether it has just arrived at that state or has been released from rest.
Question 4
Two balls are launched vertically upwards from the same height, but along separate parallel tracks.
Ball A is launched at \(t=0\) with an initial velocity of \(+19.6\,\mathrm{m\,s^{-1}}\).
Ball B is launched one second later with an initial velocity of \(+24.5\,\mathrm{m\,s^{-1}}\).
Both experience a constant downward acceleration of \(9.8\,\mathrm{m\,s^{-2}}\). Ignore air resistance.
a) Show that both balls are at the same height at \(t=2.0\) seconds, measured from the launch of Ball A.
b) Calculate the velocity of each ball at this instant.
c) Calculate the vertical separation between the balls \(0.50\) seconds later.
d) Explain why having the same height and acceleration does not mean that the balls have identical subsequent motion.
Solution 4
At \(t=2.0\) seconds, both balls are at the same height, but Ball A has zero velocity while Ball B is still moving upwards at \(14.7\,\mathrm{m\,s^{-1}}\). Half a second later, Ball B is 7.35 metres higher than Ball A.
Step 1: Calculate the position of Ball A at \(t=2.0\) seconds.
For Ball A:
\[ u_A=19.6\,\mathrm{m\,s^{-1}} \]
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
Using:
\[ \Delta y_A=u_At+\frac12at^2 \]
\[ \Delta y_A=19.6(2.0)-4.9(2.0)^2 \]
\[ \Delta y_A=39.2-19.6 \]
\[ \boxed{\Delta y_A=19.6\,\mathrm{m}} \]
Step 2: Calculate the position of Ball B at the same time.
Ball B was launched at \(t=1.0\) second.
Its elapsed time is therefore:
\[ t_B=2.0-1.0=1.0\,\mathrm{s} \]
Using the displacement equation:
\[ \Delta y_B=24.5(1.0)-4.9(1.0)^2 \]
\[ \boxed{\Delta y_B=19.6\,\mathrm{m}} \]
Both balls are at the same height.
Step 3: Calculate their velocities.
For Ball A:
\[ v_A=u_A+at \]
\[ v_A=19.6-9.8(2.0) \]
\[ \boxed{v_A=0\,\mathrm{m\,s^{-1}}} \]
Ball A is at its highest point.
For Ball B:
\[ v_B=u_B+at_B \]
\[ v_B=24.5-9.8(1.0) \]
\[ \boxed{v_B=+14.7\,\mathrm{m\,s^{-1}}} \]
Ball B is still moving upwards.
Step 4: Calculate their positions after another \(0.50\) seconds.
For Ball A, its initial velocity at \(t=2.0\) seconds is zero.
Its additional displacement is:
\[ \Delta y_A=0(0.50)-4.9(0.50)^2 \]
\[ \Delta y_A=-1.225\,\mathrm{m} \]
For Ball B, its initial velocity at \(t=2.0\) seconds is \(+14.7\,\mathrm{m\,s^{-1}}\).
Its additional displacement is:
\[ \Delta y_B=14.7(0.50)-4.9(0.50)^2 \]
\[ \Delta y_B=7.35-1.225 \]
\[ \Delta y_B=6.125\,\mathrm{m} \]
The vertical separation is:
\[ \text{Separation}=6.125-(-1.225) \]
\[ \boxed{\text{Separation}=7.35\,\mathrm{m}} \]
Ball B is now 7.35 metres higher than Ball A.
Step 5: Explain why their motions differ.
A tempting prediction is that because both balls have the same height and acceleration at \(t=2.0\) seconds, they should continue moving together.
However, their velocities are different.
Ball A has zero velocity, while Ball B is moving upwards.
Although both balls experience the same downward acceleration, their different velocities cause their positions to change differently.
Compare this with Question 3:
- In Question 3, both balls had the same position, velocity, and acceleration. Their subsequent motion was identical.
- Here, the balls have the same height and acceleration, but different velocities. Their subsequent motion is different.
Having the same acceleration alone is not enough to guarantee identical motion.
Question 5
A train travels east along a straight track at a constant velocity of \(8.0\,\mathrm{m\,s^{-1}}\).
A passenger throws a ball vertically upwards relative to the train, giving it an initial vertical velocity of \(+14.7\,\mathrm{m\,s^{-1}}\).
Assume the train maintains constant velocity and air resistance is negligible.
Consider two observers:
- Observer A is a passenger inside the train.
- Observer B is standing on the platform.
a) Calculate the time at which Observer A measures the ball’s velocity as zero.
b) Calculate the ball’s velocity relative to the ground at this instant.
c) Determine the acceleration measured by each observer.
d) A student claims that the ball cannot have zero velocity because it is still travelling east. Explain why this claim is incomplete.
Solution 5
The ball has zero velocity relative to the train at its highest point, but it still has an eastward velocity of \(8.0\,\mathrm{m\,s^{-1}}\) relative to the ground. Both observers measure a downward acceleration of \(9.8\,\mathrm{m\,s^{-2}}\).
This question introduces an important idea: velocity depends on the reference frame used to measure it.
Step 1: Calculate when the passenger measures zero velocity.
Relative to the train, the ball is moving vertically upwards with:
\[ u=+14.7\,\mathrm{m\,s^{-1}} \]
Its vertical acceleration is:
\[ a=-9.8\,\mathrm{m\,s^{-2}} \]
At the highest point, its vertical velocity is zero.
Using:
\[ v=u+at \]
\[ 0=14.7-9.8t \]
\[ \boxed{t=1.50\,\mathrm{s}} \]
The passenger observes the ball momentarily at rest after 1.50 seconds.
Step 2: Calculate its velocity relative to the ground.
The train travels east at a constant \(8.0\,\mathrm{m\,s^{-1}}\).
Because the ball is thrown vertically relative to the train, it retains the train’s horizontal velocity relative to the ground.
At the highest point, its vertical velocity is zero.
Its ground-relative velocity is therefore:
\[ \boxed{v_{\text{ground}}=8.0\,\mathrm{m\,s^{-1}}\text{ east}} \]
The platform observer sees the ball continuing eastwards, even though its vertical velocity is zero.
Step 3: Determine the acceleration measured by each observer.
For the passenger, the ball has zero horizontal velocity relative to the train throughout the flight.
Its vertical acceleration is downwards:
\[ \boxed{a_{\text{train}}=9.8\,\mathrm{m\,s^{-2}}\text{ downwards}} \]
The train moves at constant velocity relative to the ground.
Therefore, the ground and train reference frames differ by a constant velocity, not an acceleration.
Both observers measure the same acceleration:
\[ \boxed{a_{\text{ground}}=9.8\,\mathrm{m\,s^{-2}}\text{ downwards}} \]
Step 4: Evaluate the student’s claim.
The student is correct that the ball is still travelling east relative to the ground.
However, the student has assumed that velocity is independent of the observer.
It isn’t.
Observer A measures velocity relative to the train. Observer B measures velocity relative to the ground.
At the highest point:
| Quantity | Passenger’s measurement | Platform observer’s measurement |
|---|---|---|
| Horizontal velocity | Zero | \(8.0\,\mathrm{m\,s^{-1}}\) east |
| Vertical velocity | Zero | Zero |
| Total velocity | Zero | \(8.0\,\mathrm{m\,s^{-1}}\) east |
| Acceleration | \(9.8\,\mathrm{m\,s^{-2}}\) down | \(9.8\,\mathrm{m\,s^{-2}}\) down |
Both observers are correct.
The ball has zero velocity and nonzero acceleration in the passenger’s reference frame. It has nonzero velocity and nonzero acceleration in the platform observer’s reference frame.
There is no contradiction.
This final example adds one more layer to the original question: zero velocity is meaningful only when we specify the reference frame.
07The key idea to remember
An object can absolutely have zero velocity and nonzero acceleration.
The most familiar example is a ball at the highest point of its vertical flight. Its velocity is momentarily zero, but gravity continues accelerating it downwards.
When answering an HSC Physics question, remember these distinctions:
- Zero velocity means the object’s instantaneous velocity is zero. It does not automatically mean zero acceleration.
- Acceleration describes how velocity changes, not how fast an object is moving.
- The signs of velocity and acceleration depend on the chosen positive direction.
- For uniform acceleration, the velocity-zero time is found by setting \(v=0\) in \(v=u+at\).
- A mathematical root must be checked against the physical time interval and the quantity being calculated.
- Velocity must always be considered relative to a specified reference frame.
The next useful step is understanding how acceleration changes velocity over an entire time interval, rather than at just one instant. This is where interpreting acceleration-time graphs and calculating the area under them becomes important.