Acceleration Along a Straight Line Revision Notes for HSC SSCE Physics

Acceleration Along a Straight Line

What is acceleration?

When you watch sprinters in a 100 m race, they speed up very quickly at the start, then slow down after crossing the finish line. In everyday language, we might call the first part "acceleration" and the second part "deceleration." However, in physics, both are actually examples of acceleration – one is positive acceleration (speeding up) and the other is negative acceleration (slowing down).

infoNote

In physics, we use the term acceleration for any change in velocity, whether speeding up or slowing down. This differs from everyday language where "acceleration" typically means only speeding up.

Acceleration measures how quickly velocity changes over a period of time. It occurs whenever there is a change in velocity, which includes:

Changes in speed (getting faster or slower)
Changes in direction (even if speed stays constant)

For motion in a straight line, which is our focus here, acceleration depends only on changes in speed, not direction.

Understanding positive and negative acceleration

The sign of acceleration depends on the direction of motion:

Positive acceleration: When speed in the positive direction is increasing
Negative acceleration: When speed in the positive direction is decreasing

chatImportant

A car going faster in reverse (negative direction) would have negative acceleration, because it's speeding up in the negative direction. The sign of acceleration depends on both the direction of motion and whether the object is speeding up or slowing down.

Calculating average acceleration

Acceleration occurs over a time interval. To find the average acceleration, we use this formula:

$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t} \quad \text{(Units: m} \cdot \text{s}^{-2}\text{)}$

Where:

$a_{\text{avg}}$ = average acceleration (metres per second squared)
$\Delta v$ = change in velocity (metres per second)
$v_f$ = final velocity
$v_i$ = initial velocity
$\Delta t$ = time interval (seconds)

infoNote

The units for acceleration are always $\text{m} \cdot \text{s}^{-2}$ (metres per second squared), which tells us how many metres per second the velocity changes each second.

lightbulbExample

Worked Example: Car Acceleration

Let's work through an example to see how this formula works in practice.

Problem: A car's speed increases from $5.0 \text{ m} \cdot \text{s}^{-1}$ to $15.0 \text{ m} \cdot \text{s}^{-1}$ in $4.0$ seconds. What was the car's average acceleration?

Solution:

Step 1: Extract the relevant data

Initial velocity: $v_i = 5.0 \text{ m} \cdot \text{s}^{-1}$
Final velocity: $v_f = 15.0 \text{ m} \cdot \text{s}^{-1}$
Time interval: $\Delta t = 4.0 \text{ s}$

Step 2: Use the acceleration formula $a_{\text{avg}} = \frac{\Delta v}{\Delta t}$

Step 3: Substitute the values $a_{\text{avg}} = \frac{v_f - v_i}{\Delta t} = \frac{15.0 \text{ m} \cdot \text{s}^{-1} - 5.0 \text{ m} \cdot \text{s}^{-1}}{4.0 \text{ s}}$

Step 4: Calculate the answer $a_{\text{avg}} = \frac{10.0 \text{ m} \cdot \text{s}^{-1}}{4.0 \text{ s}} = 2.5 \text{ m} \cdot \text{s}^{-2}$

The car's average acceleration was $2.5 \text{ m} \cdot \text{s}^{-2}$ .

Instantaneous acceleration

Just as we can find instantaneous speed at a specific moment in time, we can also find instantaneous acceleration. This is the acceleration at one exact moment, rather than averaged over a time interval.

To find instantaneous acceleration:

Make the time interval ( $\Delta t$ ) extremely small, approaching zero
Find the gradient of the velocity-time graph at that specific moment

infoNote

For motion with constant acceleration, the average acceleration equals the instantaneous acceleration: $a_{\text{avg}} = a_{\text{inst}} = a$

This is a special case that simplifies calculations significantly.

Acceleration-time graphs

Acceleration-time graphs are powerful tools for understanding motion. An important property of these graphs is that the area under an acceleration-time graph gives the change in speed.

The diagram above shows three different scenarios:

Graph a: Constant acceleration

The acceleration stays the same over time
The shaded rectangular area represents the change in speed
Area = acceleration × time interval

Graph b: Linearly increasing acceleration

The acceleration increases steadily over time
The shaded trapezoidal area represents the change in speed

Graph c: Variable acceleration

The acceleration changes in a complex way over time
Even with this irregular pattern, the shaded area still represents the change in speed

chatImportant

This relationship works for any acceleration pattern – constant, linear, or complex curves. The area under the acceleration-time graph always gives the change in speed, regardless of how complicated the motion is.

Working with velocity-time graphs

Velocity-time graphs are extremely useful for analyzing motion. You can extract two key pieces of information from these graphs:

Acceleration = gradient (slope) of the line
Displacement = area under the curve

lightbulbExample

Worked Example: Cruise Ship Motion

Let's work through a detailed example involving both types of graphs.

Problem: A cruise ship accelerates at a constant rate for $10.0$ minutes until it reaches a speed of $10 \text{ m} \cdot \text{s}^{-1}$ . It then continues in a straight line for $20.0$ minutes at $10 \text{ m} \cdot \text{s}^{-1}$ .

Sketch a speed versus time graph (converting minutes to seconds, so time goes from $0$ to $1800$ s)
Find the ship's acceleration during the first $10.0$ minutes
Sketch an acceleration versus time graph

Solution:

Part 1: Speed-time graph

The graph shows two sections:

From $0$ to $600$ s: the speed increases linearly from $0$ to $10 \text{ m} \cdot \text{s}^{-1}$
From $600$ to $1800$ s: the speed remains constant at $10 \text{ m} \cdot \text{s}^{-1}$

Note: We convert $10.0$ minutes to seconds: $10.0 \times 60 = 600$ s

Part 2: Calculate acceleration

Using the acceleration formula: $a = \frac{\Delta v}{\Delta t} = \frac{10 \text{ m} \cdot \text{s}^{-1} - 0 \text{ m} \cdot \text{s}^{-1}}{10 \text{ min} \times 60 \text{ s per min}}$

$a = \frac{10 \text{ m} \cdot \text{s}^{-1}}{600 \text{ s}} = 1.67 \times 10^{-2} \text{ m} \cdot \text{s}^{-2}$

Part 3: Acceleration-time graph

The acceleration-time graph shows:

From $0$ to $600$ s: constant acceleration of $1.67 \times 10^{-2} \text{ m} \cdot \text{s}^{-2}$
From $600$ to $1800$ s: zero acceleration (ship moving at constant speed)

lightbulbExample

Worked Example: Complex Velocity-Time Motion

Now let's analyze a more complex motion pattern.

Problem: The velocity-time graph above shows an object moving along a straight line. Find:

The displacement after $20$ s
The acceleration at $17.0$ s
The acceleration at $t = 35.0$ s

Solution:

Part 1: Finding displacement

Displacement equals the area under the velocity-time graph. We need to break this into sections (triangles and rectangles):

\begin{align} s &= \frac{1}{2}(10 \text{ m} \cdot \text{s}^{-1} \times 10 \text{ s}) + (10 \text{ m} \cdot \text{s}^{-1} \times 5 \text{ s}) \\ &\quad + \frac{1}{2}(10 \text{ m} \cdot \text{s}^{-1} \times 2.5 \text{ s}) - \frac{1}{2}(10 \text{ m} \cdot \text{s}^{-1} \times 2.5 \text{ s}) \end{align}

$s = 50 \text{ m} + 50 \text{ m} + 12.5 \text{ m} - 12.5 \text{ m} = 100 \text{ m}$

The displacement after $20$ s is $100$ m.

Part 2: Finding acceleration at $17.0$ s

At $17.0$ s, the object is on the downward sloping section. The acceleration is the change in velocity divided by time:

$a = \frac{(-10 \text{ m} \cdot \text{s}^{-1}) - (+10 \text{ m} \cdot \text{s}^{-1})}{20 \text{ s} - 15 \text{ s}} = \frac{-20 \text{ m} \cdot \text{s}^{-1}}{5 \text{ s}} = -4.0 \text{ m} \cdot \text{s}^{-2}$

The negative sign indicates the object is slowing down in the positive direction.

Part 3: Finding acceleration at $t = 35.0$ s

To find the instantaneous acceleration at this point, we find the gradient of the tangent to the curve. Looking at the graph between $t = 25$ s and $t = 40$ s:

$a = \frac{5 \text{ m} \cdot \text{s}^{-1} - 0 \text{ m} \cdot \text{s}^{-1}}{40 \text{ s} - 25 \text{ s}} = \frac{5 \text{ m} \cdot \text{s}^{-1}}{15 \text{ s}} = 0.3 \text{ m} \cdot \text{s}^{-2}$

bookmarkSummary

Exam Tips

Always include units in your calculations and final answers. Acceleration is measured in $\text{m} \cdot \text{s}^{-2}$ .
Watch your signs: Negative acceleration doesn't always mean slowing down – it depends on the direction of motion.
For velocity-time graphs:
- Gradient = acceleration
- Area = displacement
For acceleration-time graphs:
- Area = change in speed
When finding areas under curves: Break complex shapes into triangles, rectangles, and trapezoids for easier calculation.
Convert units carefully: Make sure all times are in seconds and speeds are in $\text{m} \cdot \text{s}^{-1}$ before calculating.

bookmarkSummary

Key Points to Remember

Acceleration is the rate of change of velocity, calculated as $a = \frac{\Delta v}{\Delta t}$
Acceleration can be positive (speeding up in the positive direction) or negative (slowing down in the positive direction)
Average acceleration is found over a time interval, while instantaneous acceleration is the gradient of a velocity-time graph at a specific moment
On a velocity-time graph: the gradient gives acceleration, and the area gives displacement
On an acceleration-time graph: the area gives the change in speed

Acceleration Along a Straight Line (HSC SSCE Physics): Revision Notes