Select constant acceleration formulae and keep directed signs
Choose a relation that uses the known quantities, solve constant-acceleration motion and interpret both roots in vertical or reversing journeys.
Section 4.2 · Kinematics of motion in a straight line
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Solve linear and quadratic equations.
- Distinguish signed displacement and velocity from distance and speed.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which constant-acceleration equation avoids time when u, v, a and s are involved?
What you will learn
- Select and use the appropriate constant-acceleration equation.
- Interpret signs, stopping times and vertical motion within a stated model.
Choose an equation by its assumptions and unknowns
For constant acceleration a over a time t, with initial velocity u and final velocity v, use v = u + at, s = ut + ½at², v² = u² + 2as or s = ½(u + v)t. Here s is displacement from the interval’s starting position, not absolute position or necessarily distance. All directed quantities use the same positive direction. Choose the relation that avoids an unnecessary unknown: if time is absent, v² = u² + 2as is often useful.
For u = 4 m s⁻¹, a = 3 m s⁻² and t = 2 s, v = 10 m s⁻¹ and s = 14 m. The average-velocity equation independently gives ½(4 + 10) × 2 = 14 m because velocity varies linearly. That endpoint average is not a general rule for changing acceleration. If a depends on time, use calculus or a correctly split model instead of substituting a selected instantaneous acceleration into these equations.
Vertical motion is still a signed straight-line model
An idealised particle projected vertically upwards at 20 m s⁻¹ under constant downward acceleration of magnitude 10 m s⁻² has u = +20 and a = −10 when up is positive. With no air resistance, it reaches v = 0 at 2 s and rises 20 m. It returns to its launch position at 4 s with v = −20 m s⁻¹. Gravity remains downward even at the highest point, so acceleration there is not zero.
Squaring velocity loses its sign: v² = 400 permits +20 and −20, and the motion interval decides which applies. Similarly s = 20t − 5t² = 0 has t = 0 at launch and t = 4 at return. State why a root is included or excluded; do not discard a positive time merely because a second root exists. Total distance to return is 40 m even though displacement is zero. This is one-dimensional vertical motion, not a two-dimensional projectile model.
Pause and explain
With up positive, what acceleration should be used for idealised vertical motion under g = 10 m s⁻²?
Worked example
A particle has initial velocity +4 m s⁻¹ and constant acceleration +3 m s⁻² for 2 s. Find final velocity and displacement.
Show the worked solution
- Use v = u + at to obtain v = 4 + 3 × 2 = 10 m s⁻¹.
- Use s = ut + ½at² to obtain s = 4 × 2 + ½ × 3 × 4 = 14 m.
- Check with s = ½(u + v)t = ½(4 + 10) × 2 = 14 m.
Answer Final velocity +10 m s⁻¹; displacement +14 m.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle starts with u = +5 m s⁻¹ and has constant a = +2 m s⁻² for 3 s. Find v and s and check using mean velocity.
Give me a hint
All supplied directed quantities use the same positive direction.
Compare my reasoning
- v = 5 + 2 × 3 = 11 m s⁻¹.
- s = 5 × 3 + ½ × 2 × 3² = 24 m.
- The constant-acceleration mean gives ½(5 + 11) × 3 = 24 m again.
v = +11 m s⁻¹; s = +24 m.
Look for these in your work
- Use the squared elapsed time in the acceleration term.
- Confirm the displacement with the endpoint average valid for this model.
A particle travelling positively at 18 m s⁻¹ stops after displacement +54 m under constant acceleration. Find acceleration and stopping time.
Give me a hint
Use the relation without time first, then the velocity equation.
Compare my reasoning
- With v = 0, 0² = 18² + 2a × 54 gives a = −3 m s⁻².
- Then 0 = 18 − 3t gives t = 6 s.
- The mean-velocity check gives ½(18 + 0) × 6 = 54 m.
Acceleration −3 m s⁻²; stopping time 6 s.
Look for these in your work
- Use zero final velocity and retain the directed acceleration sign.
- Check that time is positive and the stopping displacement agrees.
A particle is projected vertically up at 20 m s⁻¹ from a launch point. Ignore air resistance and use g = 10 m s⁻². Find maximum height above launch, time to return, return velocity and total distance to return.
Give me a hint
Choose up as positive and distinguish launch from the later zero-displacement root.
Compare my reasoning
- At the top, 0 = 20 − 10t gives t = 2 s; height is 20 × 2 − 5 × 2² = 20 m.
- Return requires 20t − 5t² = 0: discard launch t = 0 for the later event and obtain t = 4 s.
- Return velocity is 20 − 10 × 4 = −20 m s⁻¹; total distance is 20 + 20 = 40 m although displacement is zero.
Maximum height 20 m; return time 4 s; return velocity −20 m s⁻¹; total distance 40 m.
Look for these in your work
- Keep a = −10 on both rising and falling intervals.
- Explain the time root and distinguish total distance from zero net displacement.
Common mistakes
- Check acceleration is constant on the whole chosen interval.
- The squared-velocity equation does not determine a velocity sign.
- Acceleration at the top of vertical motion remains downward.
What can you explain now?
A particle projected upwards is momentarily at rest at its highest point. Is its gravitational acceleration zero?
Compare with the explanation
No; it remains downward with magnitude g in this model.
Zero velocity and zero acceleration are different statements.
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Make it stick
Check constant acceleration, choose a direction and an equation that fits the data, then interpret signs and event roots.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.