Vectors and scalars, the equations of motion for constant acceleration, and the interpretation of velocity-time and acceleration-time graphs for motion in a straight line.

What this key area is asking

The SQA wants you to distinguish vectors from scalars, select and use the four equations of motion for constant acceleration, understand where they come from, and interpret velocity-time and acceleration-time graphs, including finding displacement from the area under a graph and acceleration from a gradient.

Vectors and scalars

Displacement is the straight-line distance from start to finish in a stated direction, and velocity is the rate of change of displacement, so both are vectors. Distance and speed are the corresponding scalars. To add vectors that are not in a straight line, draw them tip-to-tail and use Pythagoras and trigonometry on the closing side. For two perpendicular vectors of magnitude $x$ and $y$ the resultant magnitude is $R = \sqrt{x^2 + y^2}$ at an angle $\theta = \tan^{-1}\!\left(\frac{y}{x}\right)$ to the $x$ direction.

A common SQA scenario is a walker who travels $30 \text{ m}$ east then $40 \text{ m}$ north. The total distance is $70 \text{ m}$ (a scalar sum) but the displacement is $\sqrt{30^2 + 40^2} = 50 \text{ m}$ at $\tan^{-1}(40/30) = 53^{\circ}$ north of east.

The equations of motion

These apply only when the acceleration is constant and motion is in a straight line.

The first equation follows directly from the definition of acceleration, $a = \frac{v - u}{t}$, rearranged. The fourth equation is the area of a velocity-time trapezium of parallel sides $u$ and $v$ and width $t$. Substituting $v = u + at$ into $s = \frac{1}{2}(u + v)t$ gives $s = ut + \frac{1}{2}at^2$, and eliminating $t$ between the first and fourth gives $v^2 = u^2 + 2as$.

Choose the equation that contains the three quantities you know plus the one you want. A consistent sign convention is essential: pick a positive direction and make any quantity in the opposite direction negative. For an object thrown upward, the initial velocity is positive but $a = g$ is negative.

Motion graphs

For motion that changes direction, velocity becomes negative, so the area below the time axis counts as negative displacement. This is how the SQA distinguishes total distance (add the area magnitudes) from displacement (add the signed areas). A ball bouncing on a floor traces a sawtooth velocity-time graph, with the sign flipping at each bounce.

The gradient of a displacement-time graph gives the velocity, so a curved displacement-time graph means changing velocity, and the steepness of the curve at any instant is the instantaneous speed.

Examples in context

A sprinter accelerating from the blocks reaches $9.0 \text{ m s}^{-1}$ in $2.0 \text{ s}$. The acceleration is $a = \frac{9.0 - 0}{2.0} = 4.5 \text{ m s}^{-2}$ and the distance covered in that phase is $s = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 9.0)(2.0) = 9.0 \text{ m}$.

A stone dropped down a well hits the water after $1.5 \text{ s}$. With $u = 0$ and $a = 9.8 \text{ m s}^{-2}$, the depth is $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(9.8)(1.5)^2 = 11 \text{ m}$, and the impact speed is $v = u + at = 9.8 \times 1.5 = 15 \text{ m s}^{-1}$. These data-logging style problems, often using a light gate or motion sensor, are standard SQA Higher contexts.

Try this

Q1. A car accelerates from $5\text{ m s}^{-1}$ to $20\text{ m s}^{-1}$ in $3\text{ s}$. Find its acceleration. [2 marks]

• Cue. $a = (v - u)/t = (20 - 5)/3 = 5\text{ m s}^{-2}$.

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

• Cue. The displacement.

Q3. A ball is dropped from rest and falls for $2.0\text{ s}$. Taking $g = 9.8\text{ m s}^{-2}$, calculate the distance fallen. [2 marks]

• Cue. $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(9.8)(2.0)^2 = 19.6\text{ m}$.