ScienceCBSE Class 9 · Describing Motion Around Us

Equations of Motion, Free Fall, Stopping Distance and Circular Motion (Class 9)

Updated 17 October 2026 · Fact-checked

For motion in a straight line with constant acceleration, v = u + at, s = ut + ½at² and v² = u² + 2as, where u is the initial velocity, v the final velocity, a the acceleration, t the time and s the displacement. A dropped object has a = g = 9.8 m s⁻² downwards; stopping distance is reaction distance (u × reaction time) plus braking distance u² ÷ 2|a|; and in uniform circular motion the speed is constant but the velocity changes, so the motion is accelerated.

Slope of the line is a, giving v = u + at; the area under it, rectangle plus triangle, is the displacement s = ut + ½at².

Illustration drawn for QuestionBuddy.

Understand Equations of Motion, Free Fall and Stopping Distance

The three equations. For constant acceleration, take a velocity-time graph that rises in a straight line from velocity u at time 0 to velocity v at time t. The slope gives a = (v − u) ÷ t, so v = u + at. The area under the graph is the displacement: a rectangle of height u and width t, plus a triangle of height (v − u) and base t, so s = ut + ½(v − u)t. Since v − u = at, s = ut + ½at². Putting t = (v − u) ÷ a into this gives v² = u² + 2as. Two more forms are s = ½(u + v)t (area of a trapezium) and s = vt − ½at². These hold only when acceleration is constant, and all quantities must carry signs: if the positive direction is the direction of motion, a slowing object has negative a.

Choosing the equation. List u, v, a, s, t with what is given and what is wanted; use the equation that has the unknown and does not need the quantity you do not know. If t is not given, use v² = u² + 2as. If v is not given, use s = ut + ½at². If s is not given, use v = u + at. Convert km h⁻¹ to m s⁻¹ first (multiply by 5/18).

Free fall. An object dropped from rest gains 9.8 m s⁻¹ of speed every second, so after 1, 2, 3 and 4 s its speed is 9.8, 19.6, 29.4 and 39.2 m s⁻¹. This constant acceleration is the acceleration due to gravity, g = 9.8 m s⁻², directed downwards. For an object dropped from rest, v = gt and s = ½gt². For an object thrown upwards, take up as positive, so a = −g: at the top, v = 0, the height is u² ÷ 2g and the time to the top is u ÷ g; the time to return to the starting level is twice that. The mass of the object does not appear in the equations (air resistance neglected).

Stopping distance and road safety. When the driver sees danger, the vehicle first travels a reaction distance at constant speed, u × t_reaction, and then a braking distance while it slows to rest. From 0 = u² + 2as, the braking distance is s = u² ÷ 2|a|. Stopping distance = reaction distance + braking distance. Doubling the speed makes the braking distance four times as long. Braking distance is longer on wet roads, with worn tyres or with weak brakes (smaller |a|), and reaction distance is longer if the driver is tired or distracted.

Uniform circular motion. For one revolution of a circle of radius R, distance = 2πR and displacement = 0. Average speed = 2πR ÷ T, and average velocity over a whole revolution is zero. Uniform circular motion means constant speed on a circle; the velocity at any point is along the tangent, and since its direction keeps changing, the velocity changes and the motion is accelerated. A marble leaving a ring moves off along the tangent.

Equations of Motion, Free Fall and Stopping Distance: formulas and key facts

First equation of motion: $v = u + a t$
Second equation of motion: $s = u t + ½ a t^{2}$
Third equation of motion: $v^{2} = u^{2} + 2 a s$
Average-velocity form: $s = ½ (u + v) t$
Free fall (from rest): $g = 9.8 m s^{- 2} d o w n w a r d s ∣ v = g t ∣ s = ½ g t^{2}$
Stopping distance: $R e a c t i o n d i s t an ce u \times t_{r} e a c t i o n + b r ak in g d i s t an ce u^{2} \div 2∣ a ∣$
Uniform circular motion: $v = 2 π R \div T ∣ d i s pl a ce m e n t a f t er o n er e v o l u t i o n = 0$

How to solve Equations of Motion, Free Fall and Stopping Distance questions

Use this method for any constant-acceleration numerical.

1Draw a quick sketch, choose a positive direction and convert all units to m and s.
2List u, v, a, s and t, writing '?' for the unknown and noting values such as 'from rest' (u = 0) or 'comes to rest' (v = 0).
3Pick the equation that contains the unknown and the knowns but not the missing quantity.
4Substitute with signs and units, solve, and check that the sign and size are reasonable.
5For a multi-stage problem, treat each stage separately and carry the final velocity of one stage as the initial velocity of the next.
6Check by using a second equation or by counting the area under the velocity-time graph.

Equations of Motion, Free Fall and Stopping Distance in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Can a scooter stop in time?

Suppose a scooter rider travels at 36 km h⁻¹ with a reaction time of 0.6 s, and her brakes give a constant deceleration of 5 m s⁻². A child steps onto the road 14 m ahead.

Find the stopping distance and decide whether she can stop before reaching the child.

Try it first, then show the solutionSolution

36 km h⁻¹ = 36 × 5/18 = 10 m s⁻¹.
Reaction distance = 10 × 0.6 = 6 m. Braking distance = 10² ÷ (2 × 5) = 100 ÷ 10 = 10 m.
Stopping distance = 6 + 10 = 16 m, which is 2 m more than the 14 m available, so she cannot stop in time.

Answer: 16 m; no, she would be about 2 m short.

Memory hook: Add the thinking distance to the braking distance.

Dropping a stone to estimate a height

Suppose a student drops a pebble from a bridge and a stopwatch shows that it takes 2.0 s to hit the water.

Estimate the height of the bridge and the speed of the pebble as it hits the water (g = 9.8 m s⁻², ignore air resistance).

Try it first, then show the solutionSolution

Height: s = ½gt² = ½ × 9.8 × 2.0² = ½ × 9.8 × 4 = 19.6 m.
Speed: v = gt = 9.8 × 2.0 = 19.6 m s⁻¹.
Check: v² = 2gs = 2 × 9.8 × 19.6 = 384.16, and 19.6² = 384.16.

Answer: 19.6 m and 19.6 m s⁻¹.

Memory hook: Starts from rest: v = gt and s = ½gt².

Tricks to remember Equations of Motion, Free Fall and Stopping Distance

Short memory aids. Say each one aloud once, then use it on a question the same day.

Three equations, three missing letters
v = u + at has no s. s = ut + ½at² has no v. v² = u² + 2as has no t. Pick the one that misses the letter you do not know.
Use it when: Choosing the right equation fast.
Up is positive, g is negative
For an upward throw, a = −9.8 m s⁻². At the top v = 0, and the way up takes as long as the way down.
Use it when: Projectile thrown straight up.
Square the speed
Braking distance ∝ u². Twice the speed, four times the braking distance; three times the speed, nine times.
Use it when: Stopping distance reasoning.
Circle: speed same, velocity changes
Every point has a new tangent direction, so a velocity change happens at each point.
Use it when: Why circular motion is accelerated.

Equations of Motion, Free Fall and Stopping Distance: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A car accelerating from rest · 3 marks · Short answerBoard level

A car starts from rest and moves with a constant acceleration of 2 m s⁻² for 8 s. Find (a) its final velocity, (b) the distance covered in 8 s, and (c) check the distance with v² = u² + 2as.

Try it first, then show the solutionSolution

(a) u = 0, a = 2 m s⁻², t = 8 s. v = u + at = 0 + 2 × 8 = 16 m s⁻¹. (1)
(b) s = ut + ½at² = 0 + ½ × 2 × 8² = 64 m. (1)
(c) v² = 16² = 256 and 2as = 2 × 2 × 64 = 256, so s = 64 m is consistent. (1)

Answer: (a) 16 m s⁻¹; (b) 64 m; (c) 256 = 256, consistent.

Example 2: A braking car · 3 marks · Short answerBoard level

A car moving at 20 m s⁻¹ is brought to rest by brakes giving a constant retardation of 4 m s⁻². Find (a) the time to stop and (b) the distance covered while braking.

Try it first, then show the solutionSolution

(a) u = 20 m s⁻¹, v = 0, a = −4 m s⁻². t = (v − u) ÷ a = (0 − 20) ÷ (−4) = 5 s. (1)
(b) v² = u² + 2as gives 0 = 400 + 2(−4)s, so s = 400 ÷ 8 = 50 m. (1)
Check: s = ½(u + v)t = ½ × 20 × 5 = 50 m. (1)

Answer: (a) 5 s; (b) 50 m.

Example 3: Free fall and a stone thrown upwards · 5 marks · Long answerHarder board level

(Take g = 9.8 m s⁻².) A ball is dropped from the top of a tower and takes 3 s to reach the ground.
(a) Find the speed on reaching the ground. (1)
(b) Find the height of the tower. (1)
(c) Find the distance fallen during the last second. (1)
A second stone is thrown vertically upwards at 19.6 m s⁻¹ from the ground.
(d) Find the maximum height and the total time it is in the air. (2)

Try it first, then show the solutionSolution

(a) v = gt = 9.8 × 3 = 29.4 m s⁻¹. (1)
(b) h = ½gt² = ½ × 9.8 × 9 = 44.1 m. (1)
(c) Distance in the first 2 s = ½ × 9.8 × 4 = 19.6 m, so the last second covers 44.1 − 19.6 = 24.5 m. (1)
(d) Up positive, a = −9.8. At the top v = 0: 0 = 19.6² − 2 × 9.8 × H, so H = 384.16 ÷ 19.6 = 19.6 m. Time to top = u ÷ g = 19.6 ÷ 9.8 = 2 s, and the time back down is also 2 s, so total time = 4 s. (2)

Answer: (a) 29.4 m s⁻¹; (b) 44.1 m; (c) 24.5 m; (d) 19.6 m and 4 s.

Example 4: Stopping distance and the wet road · 5 marks · Long answerHarder board level

A car travels at 72 km h⁻¹. The driver's reaction time is 0.5 s. On a dry road the brakes give a retardation of 4 m s⁻²; on a wet road the retardation is 2 m s⁻².
(a) Convert the speed to m s⁻¹. (1)
(b) Find the stopping distance on the dry road. (2)
(c) Find the stopping distance on the wet road. (1)
(d) On the wet road an obstacle appears 85 m ahead. At what speed does the car hit it? (1)

Try it first, then show the solutionSolution

(a) 72 × 5/18 = 20 m s⁻¹. (1)
(b) Reaction distance = 20 × 0.5 = 10 m. Braking distance = 20² ÷ (2 × 4) = 400 ÷ 8 = 50 m. Stopping distance = 10 + 50 = 60 m. (2)
(c) Braking distance on a wet road = 400 ÷ (2 × 2) = 100 m. Stopping distance = 10 + 100 = 110 m. (1)
(d) The stopping distance 110 m is more than 85 m, so the car hits it. After the 10 m reaction distance, 85 − 10 = 75 m remain for braking. v² = u² + 2as = 400 + 2(−2)(75) = 400 − 300 = 100, so v = 10 m s⁻¹. (1)

Answer: (a) 20 m s⁻¹; (b) 60 m; (c) 110 m; (d) 10 m s⁻¹.

Example 5: Catching a speeding car · 4 marks · Case-basedHOTS / competency

A speeding car passes a stationary police car at a constant speed of 20 m s⁻¹. At that instant the police car starts from rest with a constant acceleration of 2 m s⁻².
(a) After how long does the police car catch the speeding car? (1)
(b) How far have they travelled by then? (1)
(c) What is the speed of the police car at that moment? (1)
(d) What is the greatest gap between the two cars before the police car catches up? (1)

Try it first, then show the solutionSolution

(a) Positions: speeding car 20t; police car ½ × 2 × t² = t². They meet when t² = 20t, so t = 20 s (t = 0 is the start). (1)
(b) Distance = 20 × 20 = 400 m. Check: 20² = 400 m. (1)
(c) v = at = 2 × 20 = 40 m s⁻¹, twice the speed of the speeding car. (1)
(d) The gap is greatest when their speeds are equal: 2t = 20, so t = 10 s. Gap = 20 × 10 − 10² = 200 − 100 = 100 m. (1)

Answer: (a) 20 s; (b) 400 m; (c) 40 m s⁻¹; (d) 100 m.

Where students lose marks in Equations of Motion, Free Fall and Stopping Distance

The slips examiners see most often, and what to write instead.

Mistake: Substituting km h⁻¹ values directly into the equations.
The acceleration is in m s⁻² so the speed must be in m s⁻¹.
Wrong working
v² = 54² − 2 × 5 × s, giving a wrong braking distance.
Right working
54 km h⁻¹ = 15 m s⁻¹, braking distance = 15² ÷ 10 = 22.5 m.
Do this: Convert at the first line.
Usually costs: 2 marks.
Mistake: Forgetting the negative sign for retardation.
'Retardation 4 m s⁻²' is written as a = +4.
Wrong working
0 = 400 + 2(4)s, so s = −50 m (a negative distance, which is impossible)
Right working
0 = 400 + 2(−4)s, so s = 50 m
Do this: A slowing object has acceleration opposite to velocity.
Usually costs: 1 to 2 marks.
Mistake: Using the equations when acceleration is not constant.
A graph with a curve or several slopes is treated as one stage.
Wrong working
Using one value of a for a whole journey with speeding up and braking.
Right working
Split the motion into stages with constant acceleration.
Do this: Check that each stage has constant a.
Usually costs: 2 marks.
Mistake: Using ½gt² for an upward throw or a throw downward with u ≠ 0.
The simple free-fall formulas are for dropping from rest.
Wrong working
Height = ½ × 9.8 × t² for a thrown stone.
Right working
Use s = ut + ½at² with the correct u and sign of a.
Do this: Write u first, and put u = 0 only if it is dropped.
Usually costs: 2 marks.
Mistake: Calling the reaction distance part of the braking distance.
Both happen after the driver sees the danger.
Wrong working
Braking distance = 12 + 22.5 = 34.5 m
Right working
Braking distance = 22.5 m; stopping distance = reaction (12) + braking (22.5) = 34.5 m
Do this: Stopping distance has two parts, only the second involves the brakes.
Usually costs: 1 mark.

Exam tips

List the five quantities u, v, a, s, t before choosing an equation; the examiner gives method marks for the list.
Write 'from rest' as u = 0 and 'comes to rest' as v = 0, and state your sign convention for g.
In derivations, draw the velocity-time graph with points A, B, C, D labelled and show the slope and area steps.
For stopping distance, calculate the reaction and braking parts separately and then add.
After solving, check one number with a second equation or the graph.

Practice questions on Equations of Motion, Free Fall and Stopping Distance

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Write the three equations of motion for uniformly accelerated motion in a straight line, naming each symbol, and state the condition under which they hold.
Hint
The condition is about the acceleration.
Answer
v = u + at, s = ut + ½at², v² = u² + 2as, where u is initial velocity, v final velocity, a acceleration, t time and s displacement (1 mark). They hold only when the acceleration is constant (1 mark).
Q2 · 3 marks · Short answerHarder board level
A cyclist starting from rest covers the first 36 m in 6 s with uniform acceleration. Find (a) the acceleration, (b) the speed at the end of 6 s, and (c) the distance covered in the next 4 s if the acceleration continues.
Hint
Use s = ut + ½at² with u = 0 first.
Answer
(a) 36 = ½ × a × 36, so a = 2 m s⁻² (1 mark). (b) v = at = 2 × 6 = 12 m s⁻¹ (1 mark). (c) Distance in 10 s = ½ × 2 × 100 = 100 m, so the next 4 s covers 100 − 36 = 64 m; check: 12 × 4 + ½ × 2 × 16 = 48 + 16 = 64 m (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): If the speed of a car is doubled, its braking distance becomes double. Reason (R): Braking distance is proportional to the square of the initial speed for a given deceleration. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Use s = u² ÷ 2|a|.
Answer
(d). R is true: s = u² ÷ 2|a|. A is false: if u doubles, u² becomes four times, so the braking distance becomes four times as long.
Q4 · 3 marks · Short answerHarder board level
A stone on a string moves in a horizontal circle of radius 14 m, completing one revolution in 20 s (take π = 22/7). Find (a) its speed, (b) the displacement after 10 s, (c) the average velocity over one revolution, and explain whether the stone is accelerating.
Hint
After half a revolution the stone is diametrically opposite its start.
Answer
(a) Circumference = 2 × 22/7 × 14 = 88 m; speed = 88 ÷ 20 = 4.4 m s⁻¹ (1 mark). (b) After 10 s it has gone half a revolution, so displacement = diameter = 28 m (1 mark). (c) Displacement over one revolution is 0, so the average velocity is 0. The stone is accelerating because the direction of its velocity changes continuously, although the speed is constant (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
Stone A is dropped from a height of 78.4 m. At the same instant, stone B is thrown vertically downwards from the same height with an initial speed u, and B reaches the ground 2 s before A. (g = 9.8 m s⁻².) (a) Find the time taken by A and its speed on reaching the ground. (b) Find u and the speed of B on landing. (c) Does the mass of the stones affect the answers? Give a reason.
Hint
For B use s = ut + ½gt² with the time B takes.
Answer
(a) 78.4 = ½ × 9.8 × t², so t² = 16 and t = 4 s; v = 9.8 × 4 = 39.2 m s⁻¹ (1 mark). (b) B takes 4 − 2 = 2 s: 78.4 = 2u + ½ × 9.8 × 4 = 2u + 19.6, so u = 58.8 ÷ 2 = 29.4 m s⁻¹; v = u + gt = 29.4 + 19.6 = 49 m s⁻¹ (2 marks). (c) No: the equations of free fall do not contain mass, because every object falls with the same acceleration g when air resistance is ignored (1 mark).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which equation has no time in it?Show answer
v² = u² + 2as.
2.What is the value of g used in this chapter?Show answer
9.8 m s⁻² downwards.
3.How does the braking distance change when speed is doubled?Show answer
It becomes four times as long.
4.What is the displacement after one full revolution in circular motion?Show answer
Zero.
5.At a point on a circular path, which line does the velocity lie along?Show answer
The tangent to the circle at that point.
6.What is the stopping distance made of?Show answer
Reaction distance plus braking distance.

Equations of Motion, Free Fall and Stopping Distance: frequently asked questions

Why is a car turning at a constant speed said to be accelerating?

Velocity includes direction, and the direction changes at every instant, so the velocity changes even though the speed does not.

Do heavier objects fall faster?

In the absence of air resistance, all objects fall with the same acceleration g = 9.8 m s⁻², so mass does not matter.

What does a negative value of s mean?

The final position is on the negative side of the starting point, in the sign convention you chose.

Can I use v = u + at for a stone thrown upwards?

Yes, with a = −g if up is positive. At the top v = 0.

Why is following distance advice based on speed?

Braking distance grows with the square of speed, so a small increase in speed needs a much bigger gap.