Movement and Position

Edexcel awards marks for stating the correct unit alongside a numerical answer. For movement and position you must use:

• mass in kilograms (kg)
• distance / displacement in metres (m)
• speed / velocity in metres per second (m/s)
• acceleration in metres per second squared (m/s²)
• force in newtons (N)
• time in seconds (s)
• gravitational field strength in newtons per kilogram (N/kg)

Mark-scheme tip. A correct numerical value with the wrong unit (e.g. 25 km/s instead of 25 m/s) is marked wrong. Convert to SI units at the START of every calculation; if a question asks for the answer in km/h, convert your m/s value at the very end (× 3.6).

Speed is a SCALAR: it has only magnitude. "30 m/s" is a complete description.

Velocity is a VECTOR: it has magnitude AND a stated direction. "30 m/s east" is a complete description.

Two cars both moving at 30 m/s but in opposite directions have the SAME speed but DIFFERENT velocities. This distinction matters when:

• Forces act on falling objects (gravity acts downward — direction matters).
• Two objects collide head-on (their velocities have opposite signs).
• A car travels around a circuit and returns to its start (it has covered a distance but its DISPLACEMENT is zero, so its average velocity is zero even though its average speed is not).

Edexcel tip. When a 4PH1 question uses the word velocity, examine the wording carefully — a direction or a sign convention is usually expected in the answer.

A distance-time graph plots distance on the y-axis and time on the x-axis.

• Horizontal line → object is stationary (zero gradient = zero speed).
• Straight sloping line → constant speed. Steeper line = faster speed.
• Curve getting steeper → accelerating (speed increasing).
• Curve flattening → decelerating (speed decreasing).

The gradient at any point equals the speed at that instant. For a curve, draw a tangent at the point of interest and calculate its gradient.

Common Edexcel question style. Multi-stage motion: a candidate is given a graph with three or four segments and asked to (a) describe the motion in each stage and (b) calculate the speed during one specific stage. Mark schemes reward clear quote of "constant speed", "stationary", "accelerating" etc. — not just numerical gradients.

A velocity-time graph plots velocity on the y-axis against time on the x-axis. Two pieces of information are encoded:

1. Gradient = ACCELERATION (spec 1.8). Steeper line = larger acceleration. A horizontal line means CONSTANT velocity (zero acceleration). A line that slopes downward means DECELERATION.

2. Area UNDER the line = DISTANCE travelled (spec 1.9). Split the area into triangles and rectangles, calculate each piece and sum them.

Worked walkthrough. A cyclist accelerates from rest at $0.80\,\text{m/s}^2$ for 10 s, then cruises at 8.0 m/s for 20 s.

• Gradient of the sloping section: $a = \dfrac{8.0 - 0}{10} = 0.80\,\text{m/s}^2$. ✓
• Distance during acceleration = triangle area $= \tfrac{1}{2} \times 10 \times 8.0 = 40\,\text{m}$.
• Distance during cruise = rectangle area $= 20 \times 8.0 = 160\,\text{m}$.
• Total distance $= 200\,\text{m}$.

Edexcel pitfall. Candidates frequently confuse the two graph types. Use this rule of thumb: if the y-axis label includes the word velocity or speed, treat as velocity-time (area = distance). If it includes distance or displacement, treat as distance-time (gradient = speed).

Three relationships describe uniform acceleration. The two on the formula sheet are:

• $a = \dfrac{v - u}{t}$ (spec 1.6) — use when you know acceleration and time.
• $v^2 = u^2 + 2as$ (spec 1.10) — use when you DO NOT know time but know distance.

Here:

• $u$ = initial velocity (m/s)
• $v$ = final velocity (m/s)
• $a$ = acceleration (m/s²)
• $s$ = distance moved (m)

When to pick which.

• "How long does it take...?" → time appears in the answer → use $a = (v-u)/t$.
• "What distance is covered...?" → distance appears in the answer → use $v^2 = u^2 + 2as$.

Worked example. A car accelerates from 12 m/s to 30 m/s over 84 m. Find the acceleration.

$a = \dfrac{v^2 - u^2}{2s} = \dfrac{30^2 - 12^2}{2 \times 84} = \dfrac{756}{168} = 4.5\,\text{m/s}^2$

Sign conventions. Both equations assume motion in ONE direction. If the object decelerates, $a$ will come out negative — keep the sign in your answer.

Spec statement 1.5 names this as one of the 4PH1 required practicals. Examiners can ask about it on both papers.

Typical method (toy car on a ramp).

1. Apparatus. Ramp, toy car, metre rule, stopwatch (or two light gates + data logger), books/blocks to vary the release height.
2. Independent variable. Release height $h$ — vary over at least 5 values from 5 cm to 30 cm.
3. Dependent variable. Time $t$ for the car to travel a fixed distance $d$ at the bottom of the ramp.
4. Calculation. Average speed at each height: $v = d/t$.
5. Reliability. Repeat each height 3 times; take the MEAN time to reduce random error.
6. Control variables. Same car, same fixed distance, same ramp surface, same release method.

Improvements often credited.

• Use light gates in place of a stopwatch to remove reaction-time error.
• Use a longer measured distance at the bottom of the ramp so any timing error has a smaller percentage effect.

Tennis-ball variant. Drop a ball from a measured height, time the fall with a stopwatch, calculate $g$ from $h = \tfrac{1}{2} g t^2$ → rearrange to $g = 2h/t^2$. Plotting $h$ vs $t^2$ gives a straight line of gradient $g/2$.