PhysicsISC Class 11 · Motion in a Straight Line

Equations of Motion: Uniform Acceleration, Free Fall and Overtaking

Updated 17 October 2026 · Fact-checked

For a body moving in a straight line with constant acceleration a, the three equations are v = u + at, s = ut + ½at² and v² = u² + 2as. The distance covered in the nth second is u + a(2n − 1)/2. For free fall, use a = g with a consistent sign convention.

Velocity-time graph with a straight sloping line from v0 to v; area S under it, with S = ((v + v0)/2) × t. — The area under the straight v-t line is the displacement S, which equals average velocity (v + v0)/2 multiplied by time t.
Picture: Noname22124 · CC0 · Wikimedia Commons

With up as positive, a = −g all the way; the stone stops at the top (v = 0) and lands with the same speed u it left with.

Illustration drawn for QuestionBuddy.

Understand Equations of Uniformly Accelerated Motion

The three equations. For constant acceleration a, with initial velocity u, final velocity v, time t and displacement s: v = u + at, s = ut + ½at², and v² = u² + 2as. They connect five quantities, so any three known values give the other two. Choose the equation that does not contain the quantity you neither know nor need.

Sign convention. Fix a positive direction first. A quantity pointing the other way is negative. A braking car has a negative a if the motion is taken positive. A stone thrown upward has u positive and a = −g if upward is positive.

Distance in the nth second. The distance covered during the nth second alone is sₙ = u + a(2n − 1)/2. It is the difference s(n) − s(n − 1). For a body released from rest it grows as 1 : 3 : 5 : ... in successive seconds.

Motion under gravity. Near the Earth's surface, a = g ≈ 9.8 m s⁻² downwards (take 10 m s⁻² only when the question says so). For a body thrown up with speed u: maximum height H = u²/2g, time to the top = u/g, time of flight = 2u/g, and it returns to the point of projection with the same speed u. At any height the speed on the way up equals the speed on the way down.

Stopping distance. The stopping distance of a vehicle is the reaction distance (u × reaction time, at constant speed) plus the braking distance u²/2a.

Meeting and overtaking. Write the position of each body as a function of the same time, equal them, and solve. If both bodies are in free fall, their relative acceleration is zero, so the relative speed is constant.

Equations of Uniformly Accelerated Motion: formulas and key facts

Equations of motion: $v = u + a t ∣ s = u t + ½ a t^{2} ∣ v^{2} = u^{2} + 2 a s$; Valid only for constant acceleration along a straight line.
Distance in the nth second: $s_{n} = u + a (2 n - 1) /2$; Equals s(n) − s(n − 1).
Vertical throw: $H = u^{2} /2 g ∣ T = 2 u / g$; H = maximum height, T = time of flight on level ground.
Stopping distance: $d = u \times t_{r} e a c t i o n + u^{2} /2 a$; Reaction distance plus braking distance.

Equations of Uniformly Accelerated Motion 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.

Stopping a goods train

Suppose a goods train moving at 36 km/h must stop in exactly 20 m before a level crossing signal, and the brakes act uniformly.

Find the required retardation and the time taken.

Try it first, then show the solutionSolution

u = 36 km/h = 10 m s⁻¹, v = 0, s = 20 m.
v² = u² + 2as → 0 = 100 + 2a(20) → a = −2.5 m s⁻².
t = (v − u)/a = (0 − 10)/(−2.5) = 4 s.

Answer: Retardation 2.5 m s⁻², time 4 s.

Memory hook: Convert km/h to m/s by dividing by 3.6 before using any equation.

Tricks to remember Equations of Uniformly Accelerated Motion

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

Pick the missing quantity
List u, v, a, s, t, tick the unknown and the one you do not have; choose the equation without the missing one.
Use it when: Selecting an equation in a few seconds.
Odd numbers for free fall
From rest, distances in successive equal intervals are in the ratio 1 : 3 : 5 : 7.
Use it when: Checking answers in dropped-object questions.

Equations of Uniformly Accelerated Motion: solved examples

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

Example 1: Distance from rest · 1 mark · MCQBoard level

A body starts from rest with a uniform acceleration of 2 m s⁻². The distance covered in 5 s is:
(a) 10 m
(b) 25 m
(c) 50 m
(d) 100 m

Try it first, then show the solutionSolution

s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m.

Answer: (b) 25 m

Example 2: Braking car · 3 marks · Short answerBoard level

A car moving at 54 km/h is brought to rest in 5 s with uniform retardation. Find the retardation and the distance travelled.

Try it first, then show the solutionSolution

u = 54/3.6 = 15 m s⁻¹, v = 0, t = 5 s. a = (0 − 15)/5 = −3 m s⁻². (1 mark)
Distance: s = u²/(2|a|) = 225/6 = 37.5 m. (1 mark)
Check: average speed 7.5 m s⁻¹ × 5 s = 37.5 m. (1 mark)

Answer: Retardation 3 m s⁻², distance 37.5 m.

Example 3: nth second · 3 marks · Short answerHarder board level

A body moving with uniform acceleration 4 m s⁻² has an initial speed of 2 m s⁻¹. Find the distance travelled in the 5th second.

Try it first, then show the solutionSolution

s₅ = u + a(2n − 1)/2 = 2 + 4 × 9/2 = 20 m. (2 marks)
Check: s(5) = 2 × 5 + ½ × 4 × 25 = 60 m; s(4) = 8 + 32 = 40 m; difference 20 m. (1 mark)

Answer: 20 m

Example 4: Last second of a fall · 4 marks · Short answerHarder board level

A heavy ball released from a high platform covers 45 m in the final second before it lands. Find the height of the platform and the total time of fall. (g = 10 m s⁻²)

Try it first, then show the solutionSolution

Distance in the nth second from rest = g(2n − 1)/2 = 5(2n − 1). (1 mark)
5(2n − 1) = 45 → n = 5, so the ball falls for 5 s. (1 mark)
h = ½gt² = ½ × 10 × 25 = 125 m. (1 mark)
Check: s(4) = 80 m, so the last second covers 125 − 80 = 45 m. (1 mark)

Answer: Time 5 s, height 125 m.

Example 5: Overtaking · 5 marks · Long answerHOTS / competency

Car A moves at a constant 20 m s⁻¹. At the instant it passes a stationary car B, car B starts from rest with uniform acceleration 2 m s⁻². Find (i) when B overtakes A, (ii) how far they have gone, (iii) the speed of B then, and (iv) the greatest gap between them before that.

Try it first, then show the solutionSolution

Equal positions: 20t = ½ × 2 × t² → t² = 20t → t = 20 s. (1 mark)
Distance = 20 × 20 = 400 m. (1 mark)
Speed of B = 2 × 20 = 40 m s⁻¹. (1 mark)
The gap is greatest when the speeds are equal: 2t = 20 → t = 10 s. (1 mark)
Gap = 20 × 10 − ½ × 2 × 100 = 200 − 100 = 100 m. (1 mark)

Answer: B overtakes at t = 20 s after 400 m with speed 40 m s⁻¹; maximum gap 100 m at t = 10 s.

Where students lose marks in Equations of Uniformly Accelerated Motion

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

Mistake: Mixing signs for upward throws
Using g = +10 with upward velocity positive gives a wrong height.
Do this: Fix the positive direction first; then a = −g for a body thrown upward.
Usually costs: 1 to 2 marks.
Mistake: Not converting km/h
Putting 54 into an equation with m and s.
Do this: Convert to m/s (divide by 3.6) at the start.
Usually costs: 1 mark.
Mistake: Using sₙ as total distance
Taking the formula for the nth second as the distance in n seconds.
Do this: sₙ is for one second only. The distance in n seconds is ut + ½at².
Usually costs: 1 to 2 marks.

Exam tips

Write the data list u, v, a, s, t with units and signs before choosing an equation.
State g (9.8 or 10 m s⁻²) as given in the question.
Check answers by finding the same quantity two ways, for example with the average speed.
For meeting problems, form one equation in t by equating positions or by using relative speed.

Practice questions on Equations of Uniformly Accelerated Motion

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

Q1 · 1 mark · MCQBoard level
A ball is thrown vertically upwards at 20 m s⁻¹. Taking g = 10 m s⁻², the maximum height is:
(a) 10 m
(b) 20 m
(c) 40 m
(d) 80 m
Hint
H = u²/2g.
Answer
(b) 20 m
Q2 · 2 marks · Short answerBoard level
A train increases its speed from 10 m s⁻¹ to 30 m s⁻¹ in 10 s with uniform acceleration. Find the acceleration and the distance covered.
Hint
Use a = (v − u)/t, then s = (u + v)t/2.
Answer
a = 2 m s⁻². s = (10 + 30)/2 × 10 = 200 m.
Q3 · 3 marks · Short answerHarder board level
A ball is thrown vertically up at 30 m s⁻¹ (g = 10 m s⁻²). Find the maximum height, the total time of flight and its speed at a height of 25 m.
Hint
Use v² = u² − 2gh for the speed.
Answer
H = 900/20 = 45 m. T = 2 × 30/10 = 6 s. At 25 m: v² = 900 − 2 × 10 × 25 = 400, so v = 20 m s⁻¹ (the same going up or down).
Q4 · 3 marks · Short answerHarder board level
A scooter with uniform acceleration travels 7 m in its 2nd second and 19 m in its 6th second. Find its initial speed and acceleration.
Hint
Write two equations from sₙ = u + a(2n − 1)/2.
Answer
7 = u + 1.5a and 19 = u + 5.5a. Subtract: 4a = 12, a = 3 m s⁻². u = 7 − 4.5 = 2.5 m s⁻¹.
Q5 · 3 marks · Short answerHarder board level
A stone is dropped from the top of a 100 m tower and, at the same instant, another is thrown up from the ground at 50 m s⁻¹ (g = 10 m s⁻²). When and where do they meet?
Hint
Both have acceleration g, so the relative speed is constant.
Answer
Relative speed 50 m s⁻¹, gap 100 m, so t = 2 s. Height of the upward stone: 50 × 2 − 5 × 4 = 80 m. Dropped stone: 100 − 20 = 80 m. They meet 80 m above the ground after 2 s.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a car moving at 30 m s⁻¹ (108 km/h) has a driver whose reaction time is 0.5 s; the brakes then give a retardation of 6 m s⁻².
(i) Find the distance moved during the reaction time. (1 mark)
(ii) Find the braking distance. (1 mark)
(iii) Find the total stopping distance. (1 mark)
(iv) If the speed were 60 m s⁻¹ instead, find the stopping distance. (1 mark)
Hint
Reaction distance is at constant speed.
Answer
(i) 30 × 0.5 = 15 m. (ii) 900/(2 × 6) = 75 m. (iii) 90 m. (iv) 60 × 0.5 = 30 m and 3600/12 = 300 m, total 330 m. Doubling the speed raises the stopping distance to about 3.7 times.

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.Equation of motion without time?Show answer
v² = u² + 2as.
2.Distance in the nth second?Show answer
u + a(2n − 1)/2.
3.Time of flight of a body thrown up at u?Show answer
2u/g.
4.Stopping distance has which two parts?Show answer
Reaction distance and braking distance.

Equations of Uniformly Accelerated Motion: frequently asked questions

Can I use these equations when acceleration changes?

No. They hold only for constant acceleration. For a variable acceleration, use calculus: integrate a to get v, and v to get x.

Why does a ball return with the same speed it was thrown?

Because the acceleration is constant and the motion is symmetrical: v² = u² − 2gh gives the same speed at each height going up and coming down, so at the starting level the speed equals u.

Is g always 9.8 m s⁻²?

Take 9.8 m s⁻² unless the question says 10 m s⁻². The page uses the value stated in each problem.