Velocity and acceleration: defining and calculating acceleration, and interpreting velocity-time graphs to describe motion and to find acceleration and distance travelled.

What this key area is asking

The SQA wants you to define acceleration, calculate it from a change in velocity, describe how it is measured, and read a velocity-time graph to describe the motion, find the acceleration from the gradient and find the distance travelled from the area under the line.

Defining and calculating acceleration

If the final velocity is less than the initial velocity, $v - u$ is negative, so the acceleration is negative. A negative acceleration means the object is slowing down (sometimes called deceleration). The sign always comes out of the numbers, so you do not add a minus sign by hand.

Measuring acceleration

In the lab, acceleration is found by measuring two velocities and the time between them. A common method uses a vehicle with a card of known length passing through two light gates connected to a timer. Each gate measures the velocity as $v = \frac{\text{length of card}}{\text{time card blocks the gate}}$, and the timer records the time to travel from one gate to the other. The acceleration is then $a = \frac{v - u}{t}$. A double mask on a single gate can also give both velocities at one point.

Reading a velocity-time graph

To find the acceleration, work out the gradient: take the change in velocity (the rise) and divide by the change in time (the run). To find the distance, find the area between the line and the time axis. Split that area into triangles and rectangles, work out each one and add them.

Describing motion in words

The SQA often asks you to describe the motion shown by a graph. Use the shape of each section: a flat line is "constant velocity"; an upward slope is "uniform (constant) acceleration, speeding up"; a downward slope towards the axis is "uniform deceleration, slowing down"; and a line reaching the time axis is "comes to rest (stops)". A steeper slope means a larger acceleration.

Try this

Q1. A train slows from $24 \text{ m s}^{-1}$ to $4.0 \text{ m s}^{-1}$ in $5.0 \text{ s}$. Calculate the acceleration. [2 marks]

• Cue. $a = (4.0 - 24)/5.0 = -4.0 \text{ m s}^{-2}$ (negative means slowing down).

Q2. State what the area under a velocity-time graph represents. [1 mark]

• Cue. The distance travelled.

Q3. A trolley accelerates uniformly from rest, reaching $6.0 \text{ m s}^{-1}$ in $3.0 \text{ s}$. Find the distance travelled. [2 marks]

• Cue. Area of triangle $= \frac{1}{2} \times 3.0 \times 6.0 = 9.0 \text{ m}$.