PhysicsISC Class 11 · Chapter 2

Motion in a Straight Line ISC Class 11: Notes and Numericals

Updated 15 October 2026 · Fact-checked

Motion in a Straight Line is Chapter 2 of ISC Class 11 Physics. It covers frames of reference, distance and displacement, average and instantaneous speed and velocity, acceleration, the elementary calculus used to describe motion, x-t, v-t and a-t graphs, the equations of uniformly accelerated motion and motion under gravity.

What Motion in a Straight Line is about

Kinematics describes motion without asking what causes it. This chapter treats the simplest case: a point mass moving along a straight line, so position needs one coordinate x measured from an origin in a chosen frame of reference.

Distance is the total path length (a scalar, never negative); displacement is the change in position (it has a sign showing direction). Average velocity = displacement/time; average speed = distance/time. Instantaneous velocity is the limit of average velocity as the time interval shrinks: $v = \frac{d x}{d t}$ , the slope of the x-t graph. Acceleration is $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$ , the slope of the v-t graph. Going the other way, displacement is the area under the v-t graph, $\int v d t$ , and change in velocity is the area under the a-t graph.

The syllabus asks for elementary differentiation (of $x^{n}$ , sin x, cos x, $e^{x}$ and ln x, as rates of change) and simple integration (mainly $\int x^{n} d x$ , with definite and indefinite integrals mentioned). With them you derive the equations of uniformly accelerated motion: $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , $v^{2} = u^{2} + 2 a s$ and $s_{n} = u + \frac{a}{2} (2 n - 1)$ , by graphical, calculus and analytical methods. Motion under gravity is the most important application: near the Earth every freely falling body has the same downward acceleration g ≈ 9.8 m s⁻², whatever its mass, if air resistance is ignored.

Weightage: Motion in a Straight Line belongs to unit 2 of the ISC Class XI Physics syllabus. CISCE weights units 1-3 together (Physical World and Measurement; Kinematics; Laws of Motion) at 23 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Frame of reference, distance and displacement: Motion is always relative: we describe it from a frame of reference, a coordinate system with an origin and a clock attached to some observer. A body treated as a point mass is one whose size is negligible compared with the distance it moves. Distance (path length) is the total length of the path travelled: a positive scalar. Displacement $Δ x = x_{2} - x_{1}$ is the change in position: it can be positive, negative or zero, and its magnitude is never more than the distance. A body that returns to its starting point has zero displacement but non-zero distance. Uniform motion: equal displacements in equal intervals of time, however small; otherwise the motion is non-uniform.; Worked example
An athlete runs one and a half laps of a 400 m circular track. Find the distance covered and the magnitude of the displacement.
Try it first, then show the solutionSolution
Distance = 1.5 × 400 = 600 m.
Circumference 2πr = 400 m, so r = 63.7 m. After 1.5 laps the athlete is diametrically opposite the start.
Displacement = 2r = 127 m.
Answer: 600 m; 127 m
Average speed and average velocity: Average speed = total distance/total time; average velocity = total displacement/total time, $\overset{v}{ˉ} = \frac{x _{2} - x _{1}}{t _{2} - t _{1}}$ , which is the slope of the chord joining two points on the x-t graph. Average speed ≥ |average velocity|, with equality only if the body never reverses. A common trap: for two equal distances covered at speeds $v_{1}$ and $v_{2}$ , the average speed is the harmonic mean $\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$ , not the arithmetic mean; for two equal time intervals it is $\frac{v _{1} + v _{2}}{2}$ .; Worked example
A car travels 30 km at 60 km h⁻¹ and the next 30 km at 40 km h⁻¹. Find its average speed.
Try it first, then show the solutionSolution
Times: 30/60 = 0.5 h and 30/40 = 0.75 h; total 1.25 h.
Average speed = 60/1.25 = 48 km h⁻¹ (check: 2 × 60 × 40/100 = 48).
Answer: 48 km h⁻¹, not 50 km h⁻¹
Differentiation as a rate of change: The derivative $\frac{d y}{d x}$ is the rate at which y changes with x: the slope of the tangent to the y-x graph. Formulae you need: $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ ; $\frac{d}{d x} (sin x) = cos x$ ; $\frac{d}{d x} (cos x) = - sin x$ ; $\frac{d}{d x} (e^{x}) = e^{x}$ ; $\frac{d}{d x} (ln x) = \frac{1}{x}$ ; the derivative of a constant is zero, and of a sum is the sum of the derivatives; a constant factor stays outside. In kinematics: instantaneous velocity $v = lim_{Δ t \to 0} \frac{Δ x}{Δ t} = \frac{d x}{d t}$ and instantaneous acceleration $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$ .; Worked example
The position of a particle is $x = 4 t^{3} - 3 t^{2} + 2$ (x in m, t in s). Find its velocity and acceleration at t = 2 s.
Try it first, then show the solutionSolution
$v = \frac{d x}{d t} = 12 t^{2} - 6 t$ ; at t = 2: v = 48 − 12 = 36 m s⁻¹
$a = \frac{d v}{d t} = 24 t - 6$ ; at t = 2: a = 42 m s⁻²
Answer: 36 m s⁻¹; 42 m s⁻² (the acceleration is not constant)
Integration: the reverse of differentiation: Integration undoes differentiation. Indefinite integral: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ (n ≠ −1), where C is a constant fixed by an initial condition. Definite integral: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , a number equal to the area under the curve between x = a and x = b. In kinematics: displacement $Δ x = \int_{t_{1}}^{t_{2}} v d t$ (area under the v-t graph) and change in velocity $Δ v = \int_{t_{1}}^{t_{2}} a d t$ (area under the a-t graph). Areas below the time axis count as negative displacement.; Worked example
The velocity of a body is $v = 3 t^{2} + 2 t$ m s⁻¹. Find its displacement between t = 0 and t = 2 s.
Try it first, then show the solutionSolution
$Δ x = \int_{0}^{2} (3 t^{2} + 2 t) d t = [t^{3} + t^{2}]_{0}^{2}$
= (8 + 4) − 0 = 12 m
Answer: 12 m
x-t, v-t and a-t graphs: x-t graph: slope = velocity. A straight line means uniform velocity; a curve bending upward means increasing velocity; a horizontal line means rest. It can never be vertical (infinite velocity) or loop back in time. v-t graph: slope = acceleration; area between the graph and the time axis = displacement. A horizontal line means uniform velocity; a straight sloping line means uniform acceleration. a-t graph: area = change in velocity. For uniform acceleration from rest: the x-t graph is a parabola, the v-t graph a straight line through the origin and the a-t graph a horizontal line.; Worked example
A car accelerates uniformly from rest to 20 m s⁻¹ in 4 s, moves at constant speed for 6 s, then stops uniformly in 5 s. Find its acceleration in each stage and the total distance.
Try it first, then show the solutionSolution
Accelerations: 20/4 = 5 m s⁻², 0, and −20/5 = −4 m s⁻².
Area under v-t: ½ × 4 × 20 + 20 × 6 + ½ × 5 × 20 = 40 + 120 + 50
Total distance = 210 m
Answer: 5 m s⁻², 0, −4 m s⁻²; 210 m
Equations of uniformly accelerated motion: For constant acceleration a, initial velocity u and displacement s in time t: (1) $v = u + a t$ ; (2) $s = u t + \frac{1}{2} a t^{2}$ ; (3) $v^{2} = u^{2} + 2 a s$ ; (4) displacement in the n-th second $s_{n} = u + \frac{a}{2} (2 n - 1)$ . Graphical derivation: the v-t graph is a straight line from (0, u) to (t, v); slope gives (1); the area of the trapezium, $\frac{1}{2} (u + v) t$ , with (1) gives (2); eliminating t gives (3). Calculus derivation: $\frac{d v}{d t} = a \Rightarrow v = u + a t$ ; $\frac{d x}{d t} = u + a t \Rightarrow s = u t + \frac{1}{2} a t^{2}$ ; $a = v \frac{d v}{d x} \Rightarrow \int_{u}^{v} v d v = \int_{0}^{s} a d x \Rightarrow v^{2} - u^{2} = 2 a s$ . They hold only when a is constant.; Worked example
A car moving at 72 km h⁻¹ brakes with a uniform retardation of 4 m s⁻². Find the stopping time, the stopping distance and the distance covered in the third second of braking.
Try it first, then show the solutionSolution
u = 72 × 5/18 = 20 m s⁻¹. Stopping time: 0 = 20 − 4t, so t = 5 s.
Stopping distance: $0 = 2 0^{2} - 2 \times 4 \times s$ , so s = 50 m.
Third second: $s_{3} = 20 - \frac{4}{2} (2 \times 3 - 1) = 20 - 10 = 10 m$
Answer: 5 s; 50 m; 10 m
Motion under gravity: Near the Earth's surface, ignoring air resistance, every body falls with the same constant acceleration g ≈ 9.8 m s⁻² directed downward, whatever its mass (free fall). Choose a sign convention and keep it: with upward positive, a = −g. Results for a body projected upward with speed u: time to the top $t = \frac{u}{g}$ ; maximum height $H = \frac{u ^{2}}{2 g}$ ; time of flight back to the launch point $\frac{2 u}{g}$ ; it returns with the same speed u. At the highest point v = 0 but a = g (still downward). For a body dropped from rest, distances fallen in successive equal intervals are in the ratio 1 : 3 : 5 : 7 … (Galileo's law of odd numbers).; Worked example
A ball is thrown vertically upward at 19.6 m s⁻¹. Find the time to reach the top, the maximum height and the total time in the air. (g = 9.8 m s⁻²)
Try it first, then show the solutionSolution
t = u/g = 19.6/9.8 = 2 s
H = u²/2g = 384.16/19.6 = 19.6 m
Total time = 2u/g = 4 s
Answer: 2 s; 19.6 m; 4 s

Physics in real life: Motion in a Straight Line

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.

Reaction time and stopping distance

Suppose a driver moving at 54 km h⁻¹ sees a child step onto the road. Her reaction time is 0.8 s and the brakes then give a uniform retardation of 5 m s⁻².

Find the total stopping distance from the moment she sees the child.

Try it first, then show the solutionSolution

u = 54 × 5/18 = 15 m s⁻¹; during reaction the car moves at constant speed: 15 × 0.8 = 12 m
Braking: $s = \frac{u ^{2}}{2 a} = \frac{225}{10} = 22.5 m$ ; total = 34.5 m

Answer: About 34.5 m, of which 12 m is covered before the brakes even act.

Memory hook: Thinking distance grows with u; braking distance with u².

Three-stage motion between metro stations

Imagine a metro train that accelerates from rest at 1 m s⁻² to 20 m s⁻¹, cruises, then decelerates at 1 m s⁻² to stop at the next station 1.2 km away.

How long does the journey take?

Try it first, then show the solutionSolution

Speeding up: t = 20 s, s = ½ × 1 × 20² = 200 m; slowing down is the same: 20 s, 200 m
Cruise distance = 1200 − 400 = 800 m at 20 m s⁻¹: 40 s; total = 20 + 40 + 20 = 80 s

Answer: 80 s (the v-t graph is a trapezium of area 1200 m).

Memory hook: Trapezium area = journey length.

Free fall and the speed of sound

Suppose you drop a stone into a dry well and hear it hit the bottom 2.0 s later. Take g = 9.8 m s⁻² and the speed of sound as 340 m s⁻¹.

Estimate the depth, first ignoring the time taken by sound, then allowing for it.

Try it first, then show the solutionSolution

Ignoring sound: $h = \frac{1}{2} \times 9.8 \times 2. 0^{2} = 19.6 m$
Sound takes about 19.6/340 ≈ 0.06 s, so the fall takes about 1.95 s: $h = 4.9 \times 1.9 5^{2} \approx 18.5 m$

Answer: About 19.6 m as a first estimate; about 18.5 m when the sound's travel time is included.

Memory hook: Sound is fast but not instant: subtract its time.

Topics in this chapter

Motion in a Straight Line formula sheet

Average velocity and speed: $\overset{v}{ˉ} = \frac{Δ x}{Δ t}, average speed = \frac{distance}{time}$; Equal distances at $v_{1}$ , $v_{2}$ : average speed $\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$ .
Instantaneous velocity and acceleration: $v = \frac{d x}{d t}, a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}} = v \frac{d v}{d x}$; Slopes of the x-t and v-t graphs.
Standard derivatives: $\frac{d}{d x} x^{n} = n x^{n - 1}; \frac{d}{d x} sin x = cos x; \frac{d}{d x} cos x = - sin x; \frac{d}{d x} e^{x} = e^{x}; \frac{d}{d x} ln x = \frac{1}{x}$; Constants differentiate to zero.
Power-rule integral: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C (n \neq = - 1), \int_{a}^{b} f d x = F (b) - F (a)$; Displacement = ∫v dt; Δv = ∫a dt.
First equation: $v = u + a t$; Constant a only.
Second equation: $s = u t + \frac{1}{2} a t^{2}$; Also $s = \frac{( u + v )}{2} t$ .
Third equation: $v^{2} = u^{2} + 2 a s$; No time involved.
Distance in the n-th second: $s_{n} = u + \frac{a}{2} (2 n - 1)$; Unit of $s_{n}$ is metre (it is a displacement in 1 s).
Vertical throw: $t_{u p} = \frac{u}{g}, H = \frac{u ^{2}}{2 g}, T = \frac{2 u}{g}$; Same speed on return to launch level.

Motion in a Straight Line: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Velocity-time graph for uniform acceleration (derivation of the equations)

'Derive the equations of motion graphically'.

Draw

Axes: velocity on the y-axis, time on the x-axis, origin O.
A straight line from point A (0, u) on the velocity axis to point B (t, v).
Drop a perpendicular BC to the time axis, and draw AD parallel to the time axis meeting BC at D.

Label

O
A (velocity u)
B (velocity v)
C (time t)
D
slope = a
rectangle OADC
triangle ADB

Watch out: Write: slope BD/AD = a gives v = u + at; area OABC = area OADC + area ADB = ut + ½at² gives s.

Graphs of uniformly accelerated motion from rest

Sketching x-t, v-t and a-t graphs, or matching a graph to a motion.

Draw

Three small graphs side by side, all against time t.
x-t: a parabola starting at the origin and curving upward.
v-t: a straight line through the origin; a-t: a horizontal line above the time axis.

Label

x (m)
v (m s⁻¹)
a (m s⁻²)
t (s)
slope of x-t = v
slope of v-t = a
area under v-t = s

Watch out: Add the free-fall case: with upward positive, the v-t graph of a ball thrown up is a straight line with slope −g crossing the time axis at the top of the flight.

How the school exam asks Motion in a Straight Line

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Definitions and distinctions	Distance and displacement; average and instantaneous velocity; uniform and non-uniform motion; frame of reference.	Define each in one line and give a contrasting example (round trip: zero displacement, non-zero distance).
Calculus numericals	Given x(t) or v(t), find v, a, the time when the body is at rest, or the displacement over an interval.	Differentiate for v and a; integrate (definite limits) for displacement; check signs for reversal.
Equations of motion	Stopping distance, n-th second distance, two-stage motion, braking problems.	List u, v, a, s, t; pick the equation without the unknown you do not need; convert km h⁻¹ to m s⁻¹ first.
Graphs	Draw or interpret x-t, v-t and a-t graphs; find distance from area and acceleration from slope.	Split the area into triangles and rectangles; areas below the axis are negative displacement but positive distance.
Motion under gravity	Throw up, drop, throw from a tower or balloon; meeting of two bodies; last-second distance.	Fix one sign convention (upward positive, a = −g) and apply it to every quantity.
Derivations	Derive v = u + at, s = ut + ½at² and v² = u² + 2as by graphical or calculus method; derive $s_{n}$ .	Draw the labelled v-t graph or start from dv/dt = a, and show every integration step with limits.

Tricks to remember Motion in a Straight Line

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

Slope down, area up
Going x → v → a, take slopes (differentiate); going a → v → x, take areas (integrate).
Use it when: Every graph and calculus question.
The missing variable
Each equation leaves one variable out: v = u + at has no s; s = ut + ½at² has no v; v² = u² + 2as has no t.
Use it when: Choosing the right equation quickly.
Odd numbers for free fall
Dropped from rest, distances in successive seconds go 1 : 3 : 5 : 7 (5 m, 15 m, 25 m, 35 m with g = 10).
Use it when: Last-second and n-th second free-fall problems.
Top of the flight
At the top, v = 0 but a = g: 'stopped, not resting'.
Use it when: Assertion-reason and conceptual items.
Harmonic for equal distances
Equal Distances → harmonic mean (2v₁v₂/(v₁ + v₂)); equal Times → arithmetic mean.
Use it when: Average-speed traps.

Motion in a Straight Line: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: What the slope of a v-t graph gives · 1 mark · MCQExam level

The slope of the velocity-time graph of a body gives its:
(a) displacement
(b) acceleration
(c) speed
(d) distance

Try it first, then show the solutionSolution

Slope = Δv/Δt = acceleration; the area under the v-t graph gives displacement.

Answer: (b)

Example 2: Zero velocity with acceleration · 1 mark · Assertion-reasonExam level

Assertion: A body can have zero velocity and yet be accelerating.
Reason: At the highest point of its flight, a ball thrown vertically upward has zero velocity but an acceleration g downward.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Gravity acts throughout the flight, so a = g even when v = 0 for an instant; the Reason is an example that proves the Assertion.

Answer: (a)

Example 3: When is the particle at rest? · 2 marks · Short answerHarder exam level

The position of a particle is $x = t^{3} - 6 t^{2} + 9 t$ (in m, t in s). Find the times at which it is momentarily at rest and its acceleration at those times. [2]

Try it first, then show the solutionSolution

$v = 3 t^{2} - 12 t + 9 = 3 (t - 1) (t - 3)$ , so v = 0 at t = 1 s and t = 3 s.
a = 6t − 12: at t = 1 s, a = −6 m s⁻²; at t = 3 s, a = +6 m s⁻².

Answer: t = 1 s (a = −6 m s⁻²) and t = 3 s (a = +6 m s⁻²)

Example 4: Equations of motion by calculus · 3 marks · Short answerHarder exam level

Starting from the definition of acceleration, derive $v = u + a t$ and $s = u t + \frac{1}{2} a t^{2}$ for a body moving with constant acceleration a. [3]

Try it first, then show the solutionSolution

$a = \frac{d v}{d t}$ , so $\int_{u}^{v} d v = \int_{0}^{t} a d t$ , giving $v - u = a t$ , i.e. $v = u + a t$ .
$\frac{d x}{d t} = v = u + a t$ , so $\int_{0}^{s} d x = \int_{0}^{t} (u + a t) d t$ .
$s = u t + \frac{1}{2} a t^{2}$ (a is constant, so it comes out of the integral).

Answer: v = u + at; s = ut + ½at²

Example 5: Last second of a fall · 3 marks · Short answerHarder exam level

A stone dropped from the top of a tower covers 25 m in the last second of its fall. Find the height of the tower. (g = 10 m s⁻²) [3]

Try it first, then show the solutionSolution

If the fall lasts t seconds, distance in the last second = ½gt² − ½g(t − 1)² = $\frac{g}{2} (2 t - 1)$ .
5(2t − 1) = 25, so t = 3 s.
h = ½ × 10 × 3² = 45 m

Answer: 45 m

Example 6: Reading a three-stage journey · 5 marks · Long answerHarder exam level

A car starts from rest and accelerates uniformly to 15 m s⁻¹ in 10 s, moves uniformly for 20 s and then decelerates uniformly to rest in 5 s.
(a) Draw the velocity-time graph. [1]
(b) Find the acceleration in each stage. [1]
(c) Find the total distance travelled. [2]
(d) Find the average speed for the journey. [1]

Try it first, then show the solutionSolution

(a) A trapezium: a line from (0, 0) to (10, 15), a horizontal line to (30, 15), a line down to (35, 0).
(b) 15/10 = 1.5 m s⁻²; 0; −15/5 = −3 m s⁻².
(c) Area = ½ × 10 × 15 + 15 × 20 + ½ × 5 × 15 = 75 + 300 + 37.5
Total distance = 412.5 m
(d) Average speed = 412.5/35 = 11.8 m s⁻¹

Answer: 1.5, 0, −3 m s⁻²; 412.5 m; 11.8 m s⁻¹

Example 7: Thrown up from a tower · 3 marks · Short answerHOTS / competency

A ball is thrown vertically upward at 10 m s⁻¹ from the top of a tower 40 m high. Find the time it takes to reach the ground and its speed on impact. (g = 10 m s⁻²) [3]

Try it first, then show the solutionSolution

Upward positive, origin at the top: s = −40 m, u = +10 m s⁻¹, a = −10 m s⁻². −40 = 10t − 5t²
t² − 2t − 8 = 0, so (t − 4)(t + 2) = 0 and t = 4 s.
v = 10 − 10 × 4 = −30 m s⁻¹ (check: v² = 100 + 2 × 10 × 40 = 900): speed 30 m s⁻¹ downward.

Answer: 4 s; 30 m s⁻¹

Example 8: Variable acceleration · 3 marks · Short answerHOTS / competency

A particle starts from rest at the origin with acceleration $a = (6 t - 4)$ m s⁻². Find when it is next momentarily at rest and where it is then. [3]

Try it first, then show the solutionSolution

$v = \int_{0}^{t} (6 t - 4) d t = 3 t^{2} - 4 t$ ; v = 0 again when t = 4/3 s.
$x = \int_{0}^{t} (3 t^{2} - 4 t) d t = t^{3} - 2 t^{2}$
At t = 4/3 s: $x = \frac{64}{27} - \frac{32}{9} = - \frac{32}{27} = - 1.19 m$

Answer: At t = 4/3 s, 1.19 m on the negative side of the origin (the equations of uniform acceleration do not apply here)

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

A sprinter's 100 m

Suppose a sprinter's tracker shows that she accelerates uniformly from rest to 10 m s⁻¹ in the first 4 s of a 100 m race and then keeps that speed to the finish.

(i) · 1 mark
Find her acceleration in the first 4 s.
Answer
Answer: 2.5 m s⁻²
a = (10 − 0)/4 = 2.5 m s⁻².
(ii) · 1 mark
How far does she run in the first 4 s?
Answer
Answer: 20 m
s = ½ × 2.5 × 4² = 20 m (or area ½ × 4 × 10).
(iii) · 1 mark
Find her finishing time.
Answer
Answer: 12 s
Remaining 80 m at 10 m s⁻¹ takes 8 s; total 4 + 8 = 12 s.
(iv) · 2 marks
Find her average speed, and describe the shape of her x-t graph.
Answer
Answer: 8.3 m s⁻¹; a parabola for 4 s, then a straight line
100/12 = 8.33 m s⁻¹. The x-t graph curves upward (parabola) while she accelerates and becomes a straight line of slope 10 m s⁻¹, tangent to the curve at t = 4 s.

A ball from a balcony

Suppose a student on a balcony 19.6 m above the ground throws a ball vertically upward at 14.7 m s⁻¹. It rises, falls past the balcony and lands on the ground below. (g = 9.8 m s⁻²)

(i) · 1 mark
How long does the ball take to reach its highest point?
Answer
Answer: 1.5 s
t = u/g = 14.7/9.8 = 1.5 s.
(ii) · 1 mark
Find the maximum height above the ground.
Answer
Answer: 30.6 m
Rise above balcony = 14.7²/19.6 = 11.0 m; 19.6 + 11.0 = 30.6 m.
(iii) · 1 mark
Find the total time before the ball hits the ground.
Answer
Answer: 4 s
−19.6 = 14.7t − 4.9t², so t² − 3t − 4 = 0 and t = 4 s.
(iv) · 2 marks
Find the velocity with which it hits the ground, and check it with $v^{2} = u^{2} + 2 a s$ .
Answer
Answer: 24.5 m s⁻¹ downward
v = 14.7 − 9.8 × 4 = −24.5 m s⁻¹; v² = 14.7² + 2 × 9.8 × 19.6 = 216.09 + 384.16 = 600.25, so |v| = 24.5 m s⁻¹.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

The displacement of a particle is $x = 3 t^{2} - t^{3}$ (in m). Find its velocity and acceleration at t = 1 s. (2 marks)

Step	Marks
$v = 6 t - 3 t^{2}$ ; at t = 1 s, v = 3 m s⁻¹	1
$a = 6 - 6 t$ ; at t = 1 s, a = 0	1

A train moving at 36 km h⁻¹ is brought to rest in 50 m by uniform braking. Find the retardation and the time taken. (3 marks)

Step	Marks
u = 36 × 5/18 = 10 m s⁻¹	1
0 = 10² − 2a(50), so a = 1 m s⁻² (retardation)	1
0 = 10 − 1 × t, so t = 10 s	1

Using a velocity-time graph, derive the three equations of motion for uniformly accelerated motion. (5 marks)

Step	Marks
Correctly labelled v-t graph: straight line from (0, u) to (t, v)	1
Slope = (v − u)/t = a, so v = u + at	1
Area = rectangle ut + triangle ½(v − u)t	1
Substitute v − u = at: s = ut + ½at²	1
s = ½(u + v)t with t = (v − u)/a gives v² = u² + 2as	1

Where students lose marks in Motion in a Straight Line

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

Mistake: Mixing signs in vertical motion
Students take g as positive for a ball going up, or take the displacement as +40 m when the ball ends 40 m below the launch point.
Wrong working
Thrown up at 10 m s⁻¹ from a 40 m tower: 40 = 10t + 5t², t = 2 s.
Right working
Upward positive: −40 = 10t − 5t², so t = 4 s.
Do this: Write 'upward +' at the start and give every quantity its sign.
Usually costs: 2-3 marks: the whole numerical goes wrong.
Mistake: Using the equations of uniform acceleration when a varies
When a = 6t − 4 or x = t³ − 6t², students still use v = u + at.
Wrong working
a = 6t − 4, so at t = 2, v = 0 + 8 × 2 = 16 m s⁻¹.
Right working
v = ∫(6t − 4)dt = 3t² − 4t; at t = 2, v = 4 m s⁻¹.
Do this: Check whether a is constant before choosing a method; if not, use calculus.
Usually costs: Full marks for the part.
Mistake: Arithmetic mean for average speed
For equal distances at different speeds, students average the speeds directly.
Wrong working
30 km at 60 km h⁻¹ and 30 km at 40 km h⁻¹: average 50 km h⁻¹.
Right working
Total distance/total time = 60/1.25 = 48 km h⁻¹.
Do this: Always compute total distance ÷ total time.
Usually costs: 1-2 marks.
Mistake: Distance from a graph that crosses the time axis
Students add signed areas and report the displacement when the question asks for distance.
Wrong working
v = 4t − t² from 0 to 6 s: area = 0, so distance = 0.
Right working
The body reverses at t = 4 s: distance = 10.67 + 10.67 = 21.3 m; displacement = 0.
Do this: Find where v = 0 and add the magnitudes of the areas on each side.
Usually costs: 1-2 marks.
Mistake: Forgetting to convert km h⁻¹
Speeds given in km h⁻¹ are substituted directly with accelerations in m s⁻².
Wrong working
u = 72, a = −4: s = 72²/8 = 648 m.
Right working
u = 72 × 5/18 = 20 m s⁻¹; s = 400/8 = 50 m.
Do this: Multiply by 5/18 before anything else.
Usually costs: 1 mark.

Practice questions on Motion in a Straight Line

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 · MCQExam level
The area under an acceleration-time graph gives:
(a) displacement
(b) change in velocity
(c) jerk
(d) distance
Hint
∫a dt.
Answer
(b)
Q2 · 1 mark · MCQExam level
The derivative of ln x with respect to x is:
(a) x
(b) ln x
(c) 1/x
(d) eˣ
Hint
Standard result.
Answer
(c)
Q3 · 1 mark · Very short answerExam level
Define instantaneous velocity.
Hint
A limit.
Answer
The limit of the average velocity as the time interval tends to zero: v = dx/dt, the slope of the x-t graph at that instant.
Q4 · 1 mark · Very short answerExam level
Give an example of motion in which the average velocity is zero but the average speed is not.
Hint
Return journey.
Answer
A ball thrown up and caught at the same point: displacement zero, distance 2H.
Q5 · 2 marks · Short answerExam level
Differentiate $y = 4 x^{3} - 2 cos x + e^{x}$ with respect to x.
Hint
Term by term.
Answer
$\frac{d y}{d x} = 12 x^{2} + 2 sin x + e^{x}$
Q6 · 2 marks · Short answerHarder exam level
Evaluate $\int_{1}^{3} (3 x^{2} + 2 x) d x$ .
Hint
Power rule, then limits.
Answer
$[x^{3} + x^{2}]_{1}^{3} = 36 - 2 = 34$
Q7 · 2 marks · Short answerHarder exam level
A uniformly accelerated body covers 10 m in the 3rd second and 14 m in the 5th second. Find its initial velocity and acceleration.
Hint
$s_{n} = u + \frac{a}{2} (2 n - 1)$ .
Answer
u + 2.5a = 10 and u + 4.5a = 14, so a = 2 m s⁻² and u = 5 m s⁻¹.
Q8 · 2 marks · Short answerHarder exam level
A car stops in 40 m when braked from speed u. With the same retardation, what is the stopping distance from speed 2u?
Hint
s ∝ u².
Answer
160 m (four times as far).
Q9 · 2 marks · Short answerHarder exam level
A stone is dropped from a height of 80 m. One second later a second stone is thrown down from the same point. With what speed must it be thrown so that both reach the ground together? (g = 10 m s⁻²)
Hint
The second stone has 1 s less.
Answer
First stone: 80 = 5t², t = 4 s. Second has 3 s: 80 = 3u + 45, so u = 11.7 m s⁻¹.
Q10 · 3 marks · Short answerHarder exam level
The velocity of a particle is $v = 4 t - t^{2}$ m s⁻¹. Find its displacement and the distance it covers between t = 0 and t = 6 s.
Hint
v changes sign at t = 4 s.
Answer
x = 2t² − t³/3. x(4) = 10.67 m, x(6) = 0. Displacement = 0; distance = 10.67 + 10.67 = 21.3 m.
Q11 · 3 marks · Short answerHarder exam level
A ball is dropped from the top of a 45 m building, and at the same instant another is thrown vertically up from the ground at 30 m s⁻¹ along the same line. When and where do they meet? (g = 10 m s⁻²)
Hint
Gravity affects both equally.
Answer
Their separation closes at 30 m s⁻¹: t = 45/30 = 1.5 s. Height = 30(1.5) − 5(1.5)² = 33.75 m above the ground.
Q12 · 5 marks · Long answerHarder exam level
(a) Using $a = v \frac{d v}{d x}$ , derive $v^{2} = u^{2} + 2 a s$ . [2]
(b) Derive an expression for the distance travelled in the n-th second. [2]
(c) A body starting from rest covers 9 m in the 5th second. Find its acceleration. [1]
Hint
Integrate v dv from u to v.
Answer
(a) $\int_{u}^{v} v d v = \int_{0}^{s} a d x$ gives $\frac{v ^{2} - u ^{2}}{2} = a s$ . (b) $s_{n} = s (n) - s (n - 1) = u n + \frac{1}{2} a n^{2} - u (n - 1) - \frac{1}{2} a (n - 1)^{2} = u + \frac{a}{2} (2 n - 1)$ . (c) 9 = (a/2)(9), so a = 2 m s⁻².
Q13 · 3 marks · Short answerHOTS / competency
Show that for a body starting from rest with uniform acceleration, the distances covered in successive seconds are in the ratio 1 : 3 : 5 : 7 … Why does a dropped stone obey this rule while a stone thrown down does not?
Hint
Put u = 0 in $s_{n}$ .
Answer
With u = 0, $s_{n} = \frac{a}{2} (2 n - 1)$ , proportional to 1, 3, 5, 7 for n = 1, 2, 3, 4. A stone thrown down has u ≠ 0, so every $s_{n}$ has the extra term u and the ratio changes.

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.Can displacement be greater than distance?Show answer
No; |displacement| ≤ distance.
2.What does the slope of an x-t graph give?Show answer
Velocity.
3.What does the area under a v-t graph give?Show answer
Displacement.
4.d/dx (sin x) and d/dx (cos x)?Show answer
cos x and −sin x.
5.∫xⁿ dx?Show answer
$\frac{x ^{n + 1}}{n + 1} + C$ , n ≠ −1.
6.Distance in the n-th second?Show answer
$s_{n} = u + \frac{a}{2} (2 n - 1)$ .
7.Maximum height for a vertical throw at speed u?Show answer
$H = \frac{u ^{2}}{2 g}$ .
8.Acceleration at the top of a vertical throw?Show answer
g, downward.

Best order to study it

Learn frame of reference, point mass, distance and displacement, uniform and non-uniform motion.
Learn average speed and velocity, including the equal-distance trap.
Learn the five derivative formulae and ∫xⁿ dx; practise definite integrals.
Use calculus for instantaneous velocity and acceleration, and for displacement from v(t).
Learn to read and draw x-t, v-t and a-t graphs (slopes and areas).
Derive the equations of motion graphically and by calculus; then practise braking, n-th second and free-fall problems.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you distinguish distance and displacement with an example?
Can you find the average speed for equal distances at two speeds?
Can you differentiate xⁿ, sin x, cos x, eˣ and ln x?
Can you find displacement from v(t) using a definite integral?
Can you find distance and acceleration from a v-t graph?
Can you derive all three equations of motion graphically and by calculus?
Can you derive the n-th second formula?
Can you solve a ball-from-a-tower problem with consistent signs?

Motion in a Straight Line: frequently asked questions

How much calculus do I need for ISC Class 11 Physics?

The syllabus asks for differentiation as a rate of change, with formulae for xⁿ, sin x, cos x, eˣ and ln x, and simple ideas about integration, mainly ∫xⁿ dx, with definite and indefinite integrals mentioned. That is enough to find velocity, acceleration and displacement.

Which methods of deriving the equations of motion should I learn?

The syllabus allows graphical, calculus or analytical derivations. Learn the graphical method from the v-t graph and the calculus method from dv/dt = a; either may be asked.

Is relative velocity part of this chapter?

The ISC Class XI syllabus for this topic does not list relative velocity; it lists frames of reference, velocity, acceleration, graphs, the equations of motion and motion under gravity. Follow your school's teaching for any extra examples.

What value of g should I use?

Use the value the question gives, usually 9.8 m s⁻² or 10 m s⁻². If none is given, use 9.8 m s⁻² and say so.

Can a body have constant speed but changing velocity?

Not in a straight line without reversing; but in general yes, if its direction changes, as in uniform circular motion, which you study in Motion in a Plane.

Why are the equations of motion not valid for every motion?

They were derived assuming the acceleration is constant. If acceleration changes with time or position, you must use calculus: v = ∫a dt and x = ∫v dt.

Based on: CISCE ISC 2028 Physics syllabus (Classes XI and XII); CISCE Regulations for the ISC Examination Year 2028 (Class XI internal examination)