How to Find Relative Acceleration from Two Acceleration Vectors

Two small vehicles accelerate across a test track. Each has an acceleration magnitude of \(3.0\text{ m s}^{-2}\). Does that mean their relative acceleration is zero?

Only if their acceleration vectors point in exactly the same direction. If one accelerates east and the other west, their relative acceleration is \(6.0\text{ m s}^{-2}\).

To find the acceleration of object A relative to an observer travelling with object B, subtract B’s acceleration vector from A’s acceleration vector:

\[ \boxed{\vec a_{A/B}=\vec a_{A/G}-\vec a_{B/G}} \]

Here, G represents a stationary ground observer. Both acceleration vectors must be measured relative to this same observer.

The important part is that you subtract vectors, not just their magnitudes. Let’s see how that works and how to apply it to HSC-style questions.

01What does relative acceleration actually mean?

Imagine sitting in a car while another car travels alongside you. You can describe the other car’s acceleration in two different ways:

• A stationary observer on the roadside measures its acceleration relative to the ground.
• You measure its acceleration relative to your own car.

These measurements do not necessarily agree because your car might also be accelerating.

We need to identify three reference frames.

For example:

• \(\vec a_{A/G}\) is the acceleration of A relative to the ground.
• \(\vec a_{B/G}\) is the acceleration of B relative to the ground.
• \(\vec a_{A/B}\) is the acceleration of A as measured by an observer travelling with B.

The order matters. A relative to B means A minus B.

Where does the subtraction formula come from?

The formula follows directly from relative position.

The position of A relative to B is:

\[ \vec r_{A/B}=\vec r_{A/G}-\vec r_{B/G} \]

In words, subtract B’s ground position from A’s ground position to find where A is relative to B.

Differentiate once with respect to time, and you obtain relative velocity:

\[ \vec v_{A/B}=\vec v_{A/G}-\vec v_{B/G} \]

Differentiate again, and you obtain relative acceleration:

\[ \boxed{\vec a_{A/B}=\vec a_{A/G}-\vec a_{B/G}} \]

So relative acceleration is simply the difference between the two acceleration vectors.

This relationship assumes that the observer’s reference frame translates without rotating, so its coordinate axes remain parallel to the ground frame. Rotating reference frames require additional terms and are outside the standard HSC treatment.

Why you must subtract vectors, not magnitudes

This is where students often make mistakes.

Acceleration is a vector. It has both magnitude and direction.

Consider three situations, with A accelerating east at \(3.0\text{ m s}^{-2}\).

In the third situation, the relative acceleration vector is:

\[ \vec a_{A/B}=3\hat i-3\hat j \]

Its magnitude is:

\[ |\vec a_{A/B}|=\sqrt{3^2+(-3)^2} \]

\[ =4.24\text{ m s}^{-2} \]

Notice that the relative acceleration is not zero, even though the two objects have identical acceleration magnitudes.

There are three different quantities worth keeping separate:

Relative speed is the magnitude of relative velocity. It is not the same thing as relative acceleration.

If you want to revise the underlying distinction between vector quantities and scalar speed, see Can Two Objects Have the Same Speed but Different Velocities?.

02An experiment: measuring relative acceleration

How could we test whether subtracting two acceleration vectors actually gives the acceleration measured by a moving observer?

Consider an experiment using two low-friction air pucks.

The original measurement problem

Place two air pucks, A and B, on a horizontal air table. Air flows through tiny holes in the table, creating a thin cushion that allows the pucks to move with relatively little friction.

Suppose each puck has an acceleration sensor connected to a data logger.

The first version of the experiment records only the magnitude of each acceleration.

Both sensors display:

\[ 3.0\text{ m s}^{-2} \]

Can we conclude that the relative acceleration is zero?

No. The sensors have discarded the direction information.

The pucks might accelerate in the same direction, opposite directions, or perpendicular directions. These arrangements produce different relative accelerations despite identical sensor readings.

The original measurement method cannot distinguish between them.

Improving the apparatus

A better design uses an overhead camera and a coordinate grid beneath the air table.

Place a small visible marker on each puck. The camera records both markers at precisely the same time intervals.

A computer then tracks the coordinates of A and B relative to the same origin.

This revised arrangement addresses two important problems:

1. It records both horizontal and vertical motion, rather than just acceleration magnitudes.
2. It measures both objects in the same reference frame and at the same times.

The camera allows us to calculate each puck’s acceleration separately. We can also calculate the relative position of A with respect to B at each instant.

What should the experiment show?

If the relative acceleration formula is correct, the acceleration calculated from the relative-position measurements should agree with the difference between the two ground-frame acceleration vectors.

Consider the following illustrative data.

The camera records the relative position of A with respect to B at three equally spaced times.

East is positive \(x\), and north is positive \(y\).

These are illustrative measurements, not results from a claimed real experiment.

For equally spaced measurements, we can estimate acceleration using the second difference in position:

\[ a_x\approx \frac{x_3-2x_2+x_1}{(\Delta t)^2} \]

Here, \(\Delta t=0.20\text{ s}\).

For the horizontal direction:

\[ \begin{aligned} a_{A/B,x} &=\frac{0.64-2(0.48)+0.40}{0.20^2}\\ &=\frac{0.08}{0.04}\\ &=2.0\text{ m s}^{-2} \end{aligned} \]

For the vertical direction:

\[ \begin{aligned} a_{A/B,y} &=\frac{0.32-2(0.24)+0.20}{0.20^2}\\ &=\frac{0.04}{0.04}\\ &=1.0\text{ m s}^{-2} \end{aligned} \]

Therefore:

\[ \boxed{ \vec a_{A/B}=2.0\hat i+1.0\hat j \text{ m s}^{-2} } \]

The relative acceleration points north-east.

For constant acceleration, this second-difference calculation gives the acceleration exactly at the middle measurement time. With changing acceleration, it provides an approximation.

We can independently fit the individual position measurements and obtain the following illustrative ground-frame accelerations:

\[ \vec a_{A/G}=3.0\hat i+2.0\hat j \]

\[ \vec a_{B/G}=1.0\hat i+1.0\hat j \]

Subtracting:

\[ \begin{aligned} \vec a_{A/B} &=(3.0-1.0)\hat i\\ &\quad +(2.0-1.0)\hat j\\ &=2.0\hat i+1.0\hat j \end{aligned} \]

Both measurement methods agree.

The experiment supports the vector-subtraction model.

There are practical limitations. Camera calibration, timing errors, and position-measurement noise can affect the calculated acceleration. Taking second differences also makes measurement noise more noticeable, so fitting several successive position measurements is generally more reliable than relying on just three readings.

03How to calculate relative acceleration from two vectors

For HSC questions, you can use a consistent method every time.

Step 1: Identify the object and observer.

If the question asks for A’s acceleration relative to B, write:

\[ \vec a_{A/B} \]

Step 2: Identify the common reference frame.

Check that both supplied accelerations are measured relative to the same frame, such as the ground.

If one acceleration is measured relative to the ground and the other relative to the water, you must first convert them into a common frame.

Step 3: Choose coordinate axes.

A common choice is:

• East: positive \(x\).
• West: negative \(x\).
• North: positive \(y\).
• South: negative \(y\).

Step 4: Resolve angled vectors into components.

For a vector with magnitude \(a\) at an angle \(\theta\) above the positive horizontal axis:

\[ a_x=a\cos\theta \]

\[ a_y=a\sin\theta \]

Give each component the correct sign.

Step 5: Subtract the components.

\[ a_{A/B,x}=a_{A/G,x}-a_{B/G,x} \]

\[ a_{A/B,y}=a_{A/G,y}-a_{B/G,y} \]

Step 6: Find the magnitude and direction if required.

Use Pythagoras:

\[ |\vec a_{A/B}|= \sqrt{a_{A/B,x}^2+a_{A/B,y}^2} \]

Then use trigonometry to find the direction.

Remember that the signs of your components determine the correct quadrant.

If resolving vectors is the part you find difficult, revise Resolving Vectors into Components for HSC Physics.

04Worked examples

05Questions and solutions

Question 1

Two autonomous carts travel along a straight east-west track.

Cart A accelerates east at \(4.2\text{ m s}^{-2}\), while cart B accelerates east at \(1.7\text{ m s}^{-2}\).

Both accelerations are measured relative to the ground.

Calculate the acceleration of cart A relative to cart B.

Solution 1

Cart A accelerates east at \(2.5\text{ m s}^{-2}\) relative to cart B.

Choose east as positive.

The relative acceleration equation is:

\[ a_{A/B}=a_{A/G}-a_{B/G} \]

Substitute:

\[ \begin{aligned} a_{A/B}&=4.2-1.7\\ &=2.5\text{ m s}^{-2} \end{aligned} \]

Therefore:

\[ \boxed{2.5\text{ m s}^{-2}\text{ east}} \]

Both carts accelerate in the same direction, so their acceleration magnitudes can be subtracted directly in this particular one-dimensional situation.

Be careful: this shortcut only works because both vectors point east.

Question 2

Two aircraft accelerate in different directions.

Aircraft A has a ground-frame acceleration of \(8.0\text{ m s}^{-2}\), directed \(30^\circ\) north of west.

Aircraft B has a ground-frame acceleration of \(4.0\text{ m s}^{-2}\), directed \(60^\circ\) north of east.

An observer is travelling with aircraft B.

Calculate:

a) The acceleration vector of A relative to B.

b) The magnitude of the relative acceleration.

c) The direction of the relative acceleration, measured from west towards north.

Solution 2

The acceleration of A relative to B is approximately \(8.94\text{ m s}^{-2}\), directed \(3.4^\circ\) north of west.

The key is to resolve both vectors into horizontal and vertical components before subtracting them.

Step 1: Establish the coordinate system.

Choose east as positive \(x\) and north as positive \(y\).

Step 2: Resolve aircraft A’s acceleration.

Its horizontal component points west, so it is negative.

\[ \begin{aligned} a_{A/G,x} &=-8.0\cos30^\circ\\ &=-6.93\text{ m s}^{-2} \end{aligned} \]

Its vertical component points north:

\[ \begin{aligned} a_{A/G,y} &=8.0\sin30^\circ\\ &=4.00\text{ m s}^{-2} \end{aligned} \]

Step 3: Resolve aircraft B’s acceleration.

Both components are positive because B accelerates north-east.

\[ \begin{aligned} a_{B/G,x} &=4.0\cos60^\circ\\ &=2.00\text{ m s}^{-2} \end{aligned} \]

\[ \begin{aligned} a_{B/G,y} &=4.0\sin60^\circ\\ &=3.46\text{ m s}^{-2} \end{aligned} \]

Step 4: Calculate the relative acceleration components.

Subtract B’s components from A’s.

Horizontal:

\[ \begin{aligned} a_{A/B,x} &=-6.93-2.00\\ &=-8.93\text{ m s}^{-2} \end{aligned} \]

Vertical:

\[ \begin{aligned} a_{A/B,y} &=4.00-3.46\\ &=0.54\text{ m s}^{-2} \end{aligned} \]

Therefore:

\[ \boxed{ \vec a_{A/B} =-8.93\hat i+0.54\hat j \text{ m s}^{-2} } \]

Step 5: Calculate the magnitude.

\[ \begin{aligned} |\vec a_{A/B}| &=\sqrt{(-8.93)^2+(0.54)^2}\\ &\approx8.94\text{ m s}^{-2} \end{aligned} \]

Step 6: Calculate the direction.

\[ \begin{aligned} \theta &=\tan^{-1}\left(\frac{0.54}{8.93}\right)\\ &\approx3.4^\circ \end{aligned} \]

The horizontal component is negative, while the vertical component is positive.

Therefore, the direction is north of west.

\[ \boxed{ 8.94\text{ m s}^{-2}, \quad 3.4^\circ \text{ north of west} } \]

An important detail is that the vertical components almost cancel. This is why the relative acceleration points almost directly west, even though both aircraft have substantial northward acceleration components.

Question 3

Two autonomous vehicles, A and B, are travelling across a flat test area.

Their ground-frame acceleration vectors are identical:

\[ \vec a_{A/G} = \vec a_{B/G} = 2.0\hat i-1.0\hat j \text{ m s}^{-2} \]

At \(t=0\), their ground-frame velocities are:

\[ \vec v_{A/G}=10.0\hat i+2.0\hat j \text{ m s}^{-1} \]

\[ \vec v_{B/G}=4.0\hat i+2.0\hat j \text{ m s}^{-1} \]

Both vehicles maintain their constant acceleration vectors for the next \(5.0\text{ s}\).

a) Find the acceleration of A relative to B.

b) Determine the velocity of A relative to B at \(t=0\).

c) Determine the velocity of A relative to B after \(5.0\text{ s}\).

d) Explain whether A becomes stationary relative to B.

Solution 3

The relative acceleration is zero, but A is not stationary relative to B. A continues moving east at \(6.0\text{ m s}^{-1}\) relative to B.

Step 1: Calculate the relative acceleration.

\[ \vec a_{A/B}=\vec a_{A/G}-\vec a_{B/G} \]

Substituting:

\[ \begin{aligned} \vec a_{A/B} &=(2.0\hat i-1.0\hat j)\\ &\quad -(2.0\hat i-1.0\hat j)\\ &=0 \end{aligned} \]

Therefore:

\[ \boxed{\vec a_{A/B}=0} \]

Step 2: Calculate the initial relative velocity.

Use the relative velocity equation:

\[ \vec v_{A/B} = \vec v_{A/G}-\vec v_{B/G} \]

Substitute:

\[ \begin{aligned} \vec v_{A/B}(0) &=(10.0-4.0)\hat i\\ &\quad +(2.0-2.0)\hat j\\ &=6.0\hat i \end{aligned} \]

Therefore:

\[ \boxed{ \vec v_{A/B}(0)=6.0\hat i \text{ m s}^{-1} } \]

Step 3: Find the relative velocity after five seconds.

The relative acceleration is zero, so relative velocity remains constant.

Using:

\[ \vec v=\vec u+\vec at \]

we obtain:

\[ \begin{aligned} \vec v_{A/B}(5) &=6.0\hat i+(0)(5)\\ &=6.0\hat i\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{ \vec v_{A/B}(5) =6.0\hat i\text{ m s}^{-1} } \]

Step 4: Explain the motion.

A common mistake is to assume that zero relative acceleration means zero relative velocity.

It does not.

Zero relative acceleration means that the relative velocity is not changing.

Since A initially moves east at \(6.0\text{ m s}^{-1}\) relative to B, it continues doing so throughout the five seconds.

The two vehicles have the same acceleration, but different velocities.

Question 4

A passenger is sitting on a bus travelling east.

The bus accelerates east at a constant \(2.0\text{ m s}^{-2}\).

At \(t=0\), the passenger releases a small ball. The ball is initially at rest relative to the bus.

Assume:

• The ball experiences negligible air resistance.
• Its vertical acceleration is \(9.8\text{ m s}^{-2}\) downwards.
• The bus maintains its constant horizontal acceleration.
• East is positive \(x\), and up is positive \(y\).

Calculate:

a) The acceleration of the ball relative to the bus.

b) The ball’s displacement relative to the bus after \(1.5\text{ s}\).

Solution 4

The ball’s acceleration relative to the bus is:

\[ \boxed{ \vec a_{\text{ball/bus}} =-2.0\hat i-9.8\hat j \text{ m s}^{-2} } \]

After \(1.5\text{ s}\), the ball is \(2.25\text{ m}\) west and approximately \(11.0\text{ m}\) below its initial position relative to the bus.

Step 1: Identify the two acceleration vectors.

The ball experiences no horizontal acceleration after release, assuming negligible air resistance.

Therefore, its ground-frame acceleration is:

\[ \vec a_{\text{ball/G}}=0\hat i-9.8\hat j \]

The bus accelerates east:

\[ \vec a_{\text{bus/G}}=2.0\hat i+0\hat j \]

Step 2: Calculate the relative acceleration.

\[ \vec a_{\text{ball/bus}} = \vec a_{\text{ball/G}} - \vec a_{\text{bus/G}} \]

Substitute:

\[ \begin{aligned} \vec a_{\text{ball/bus}} &=(0-2.0)\hat i\\ &\quad +(-9.8-0)\hat j\\ &=-2.0\hat i-9.8\hat j \end{aligned} \]

The ball accelerates backwards and downwards relative to the bus.

Step 3: Establish the initial relative velocity.

The ball is released from rest relative to the bus.

Therefore:

\[ \vec u_{\text{ball/bus}}=0 \]

This is important because the ball initially has the same horizontal velocity as the bus.

Step 4: Calculate the relative displacement.

Use the constant-acceleration equation:

\[ \vec s=\vec ut+\frac12\vec at^2 \]

Since the initial relative velocity is zero:

\[ \vec s_{\text{ball/bus}} =\frac12\vec a_{\text{ball/bus}}t^2 \]

Substitute \(t=1.5\text{ s}\):

\[ \begin{aligned} \vec s_{\text{ball/bus}} &=\frac12(-2.0\hat i-9.8\hat j)(1.5)^2\\ &=-2.25\hat i-11.025\hat j \end{aligned} \]

Therefore:

\[ \boxed{ \vec s_{\text{ball/bus}} =-2.25\hat i-11.0\hat j \text{ m} } \]

The ball moves backwards because the bus accelerates forwards while the ball maintains its original horizontal velocity.

This is a useful example of why the observer’s acceleration cannot be ignored when describing motion from a moving reference frame.

Question 5

An idealised vertical test capsule contains a small ball. An observer inside the capsule measures the ball’s motion relative to the capsule.

At \(t=0\), the ball is released from rest relative to the capsule.

The capsule has the following motion:

• From \(t=0\) to \(t=1.0\text{ s}\), it accelerates upwards at \(2.0\text{ m s}^{-2}\).
• From \(t=1.0\text{ s}\) to \(t=2.0\text{ s}\), its motor disengages and it enters free fall.

After release, the ball also experiences free fall throughout.

Assume:

• Gravitational acceleration is \(9.8\text{ m s}^{-2}\) downwards.
• Air resistance is negligible.
• The ball does not contact the capsule’s walls, floor, or ceiling.
• The positive vertical direction is upwards.

A student makes the following claim:

“The ball and capsule both experience gravitational acceleration during the second second. Their relative acceleration is zero, so the ball must become stationary relative to the capsule.”

Evaluate this claim.

Calculate:

a) The ball’s relative acceleration during the first second.

b) The ball’s relative acceleration during the second second.

c) Its relative velocity at \(t=2.0\text{ s}\).

d) Its total displacement relative to the capsule at \(t=2.0\text{ s}\).

Solution 5

The student’s claim is incorrect.

The ball has zero relative acceleration during the second second, but it does not become stationary relative to the capsule. It already has a downward relative velocity from the first second, and that velocity remains constant.

This question is challenging because the observer’s acceleration changes during the motion. We must calculate the relative acceleration separately for each time interval.

Step 1: Calculate the relative acceleration during the first second.

Choose upwards as positive.

The ball is in free fall, so:

\[ a_{\text{ball/G}}=-9.8\text{ m s}^{-2} \]

The capsule accelerates upwards:

\[ a_{\text{capsule/G}}=+2.0\text{ m s}^{-2} \]

Subtract the capsule’s acceleration:

\[ \begin{aligned} a_{\text{ball/capsule}} &=-9.8-(+2.0)\\ &=-11.8\text{ m s}^{-2} \end{aligned} \]

Therefore:

\[ \boxed{ a_{\text{ball/capsule}}=-11.8\text{ m s}^{-2} } \]

The ball accelerates downwards relative to the capsule.

Step 2: Calculate the relative velocity after the first second.

Initially, the ball is at rest relative to the capsule:

\[ u_{\text{rel}}=0 \]

Use:

\[ v=u+at \]

After \(1.0\text{ s}\):

\[ \begin{aligned} v_{\text{rel}}(1) &=0+(-11.8)(1.0)\\ &=-11.8\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{ v_{\text{rel}}(1)=-11.8\text{ m s}^{-1} } \]

The ball is moving downwards at \(11.8\text{ m s}^{-1}\) relative to the capsule.

Step 3: Calculate the relative acceleration during the second second.

The capsule has now entered free fall.

Its acceleration is:

\[ a_{\text{capsule/G}}=-9.8\text{ m s}^{-2} \]

The ball is also in free fall:

\[ a_{\text{ball/G}}=-9.8\text{ m s}^{-2} \]

Therefore:

\[ \begin{aligned} a_{\text{ball/capsule}} &=-9.8-(-9.8)\\ &=0 \end{aligned} \]

So:

\[ \boxed{ a_{\text{ball/capsule}}=0 } \]

The relative acceleration is now zero.

Step 4: Calculate the relative velocity at \(t=2.0\text{ s}\).

This is the crucial part.

Zero relative acceleration means relative velocity remains constant during the second interval.

Therefore:

\[ \begin{aligned} v_{\text{rel}}(2) &=v_{\text{rel}}(1)+a_{\text{rel}}(1.0)\\ &=-11.8+(0)(1.0)\\ &=-11.8\text{ m s}^{-1} \end{aligned} \]

Thus:

\[ \boxed{ v_{\text{rel}}(2)=-11.8\text{ m s}^{-1} } \]

The ball continues moving downwards relative to the capsule.

Step 5: Calculate the total relative displacement.

First, calculate the displacement during the first second.

\[ s_1=ut+\frac12at^2 \]

\[ \begin{aligned} s_1 &=0+\frac12(-11.8)(1.0)^2\\ &=-5.9\text{ m} \end{aligned} \]

Now calculate the displacement during the second second.

The initial relative velocity for this interval is \(-11.8\text{ m s}^{-1}\), and relative acceleration is zero.

\[ \begin{aligned} s_2 &=ut+\frac12at^2\\ &=(-11.8)(1.0)+0\\ &=-11.8\text{ m} \end{aligned} \]

Add the two displacements:

\[ \begin{aligned} s_{\text{total}} &=s_1+s_2\\ &=-5.9-11.8\\ &=-17.7\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{ s_{\text{total}}=-17.7\text{ m} } \]

The ball is \(17.7\text{ m}\) below its initial position relative to the capsule after two seconds.

Why the student’s reasoning fails

The student correctly identifies that the relative acceleration becomes zero during the second second.

However, they incorrectly assume that zero acceleration means zero velocity.

The ball already acquired a downward relative velocity during the first second. When the capsule enters free fall, both objects experience the same gravitational acceleration, so their relative velocity stops changing.

It does not disappear.

This is the key distinction:

\[ \boxed{ \vec a_{A/B}=0 \quad\Rightarrow\quad \vec v_{A/B}=\text{constant} } \]

It does not necessarily mean:

\[ \vec v_{A/B}=0 \]

06Where to take this next

Once you can calculate relative acceleration, you can use it to predict how two objects move relative to one another over time.

The next step is combining relative acceleration with the constant-acceleration equations:

\[ \vec v_{A/B} = \vec u_{A/B}+\vec a_{A/B}t \]

and

\[ \vec r_{A/B} = \vec r_{A/B,0} +\vec u_{A/B}t +\frac12\vec a_{A/B}t^2 \]

These equations let you investigate whether two objects approach, separate, maintain a constant relative velocity, or eventually meet.

The important habit is to establish the observer’s reference frame before doing any calculations. Get that right, and the vector mathematics becomes much more manageable.