This MCQ module is based on: Equations of Motion, Distance-Time Graphs and Circular Motion
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Equations of Motion, Distance-Time Graphs and Circular Motion
🎓 Class 9
Science
CBSE
Theory
Ch 4 — Describing Motion Around Us
⏱ ~20 min
🌐 Language: [gtranslate]
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This assessment will be based on: Equations of Motion, Distance-Time Graphs and Circular Motion
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4.8 Graphical Representation of Motion
Graphs are powerful pictures of motion. With one quick glance at a distance–time or velocity–time graph, you can read the entire history of a moving object — when it sped up, when it stopped, even how far it travelled in any chosen interval.
Distance–Time Graph
On a distance–time graph, time runs along the x-axis and distance along the y-axis. The slope of the line gives the speed of the object.
Velocity–Time Graph
On a velocity–time graph, the slope equals acceleration, and the area under the curve equals the displacement (or distance, if the object never reverses direction). For uniform acceleration, the curve is a straight line.
🧮 Derive the Equations of Motion — Step through the v–t graph L3 Apply
The whole derivation lives inside this single trapezium. Click the slope line to derive \(v=u+at\); click the orange rectangle and the green triangle to read off \(s=ut+\tfrac{1}{2}at^2\).
Click the slope, the orange rectangle, or the green triangle above to step through the derivation.
4.9 Equations of Motion (Graphical Derivation)
Using Fig. 4.5, let \(u\) be the initial velocity at \(t=0\), \(v\) the final velocity at time \(t\), \(a\) the uniform acceleration, and \(s\) the displacement.
First Equation: \(v = u + at\)
Slope of line AB \(= a = \dfrac{v - u}{t - 0}\), which gives \(v - u = at\), i.e.
\[\boxed{\,v = u + at\,}\]
Second Equation: \(s = ut + \tfrac{1}{2}at^2\)
Displacement \(s\) = area of trapezium OABC = area of rectangle (u·t) + area of triangle (½ × t × (v−u)).
\(s = u t + \tfrac{1}{2} t (v - u) = ut + \tfrac{1}{2} t (at) = ut + \tfrac{1}{2} a t^2\).
\[\boxed{\,s = ut + \tfrac{1}{2}at^2\,}\]
Third Equation: \(v^2 = u^2 + 2as\)
Area of trapezium = ½ × (sum of parallel sides) × t = ½(u+v)t. Since \(t = (v-u)/a\):
\(s = \tfrac{(u+v)}{2} \cdot \dfrac{v-u}{a} = \dfrac{v^2 - u^2}{2a}\), giving:
\[\boxed{\,v^2 = u^2 + 2as\,}\]
Memorise: The three equations of uniformly accelerated motion are \(v = u+at\), \(s = ut+\tfrac{1}{2}at^2\), and \(v^2 = u^2 + 2as\). They are valid only when acceleration is uniform along a straight line.
Activity 4.3 — Read a v–t graph
L4 Analyse
Predict: A v–t graph is a straight line going from (0, 5 m/s) to (10 s, 25 m/s). Without computing, guess the acceleration and the displacement.
- Compute slope = acceleration = (25−5)/10 = 2 m/s².
- Compute area = trapezium = ½(5+25)(10) = 150 m.
a = 2 m/s², s = 150 m. Notice how the same graph gives two different physical answers from two different geometric features (slope and area).
4.10 Uniform Circular Motion
Imagine a stone tied to a string, whirled in a horizontal circle at constant speed. Although the speed is unchanging, the direction of motion changes at every instant. We say the stone is in uniform circular motion.
Speed of an object on a circular path of radius \(r\), completing one round in time \(T\):
\[v = \dfrac{2\pi r}{T}\]
Because direction of velocity is continuously changing, the body is accelerating even when its speed is constant. This acceleration always points towards the centre of the circle and is called centripetal acceleration. (You will study its formula \(a_c = v^2/r\) in higher classes.) Releasing the string makes the stone fly off along the tangent to the circle at that instant — proof that direction was changing.
Worked Numerical Examples
Example 1 — Train using v = u + at
A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming uniform acceleration, find (a) the acceleration and (b) the distance travelled in this time.
Given: \(u = 0\), \(v = 72\) km/h \(= 20\) m/s, \(t = 5\) min \(= 300\) s.
(a) \(a = (v-u)/t = 20/300 = 0.0667\) m/s².
(b) \(s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(0.0667)(300)^2 = \tfrac{1}{2}(0.0667)(90000) = 3000\) m \(= 3\) km.
Example 2 — Car braking using v² = u² + 2as
A car moving at 36 km/h is brought to rest in 5 s. Find (a) acceleration, (b) braking distance.
\(u = 36\) km/h \(= 10\) m/s, \(v = 0\), \(t = 5\) s.
(a) \(a = (0-10)/5 = -2\) m/s².
(b) \(v^2 = u^2 + 2as \Rightarrow 0 = 100 + 2(-2)s \Rightarrow s = 25\) m.
Example 3 — Free fall
A stone is dropped from the top of a tower 80 m high. Take g = 10 m/s². Find (a) time to reach the ground, (b) velocity just before hitting.
\(u = 0\), \(a = g = 10\) m/s², \(s = 80\) m.
(a) \(s = ut + \tfrac{1}{2}gt^2 \Rightarrow 80 = 0 + 5t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4\) s.
(b) \(v = u + gt = 0 + 10 \times 4 = 40\) m/s.
Example 4 — Vertical throw
A ball is thrown vertically upward with an initial velocity of 20 m/s. Take g = 10 m/s². Find (a) the maximum height reached, (b) the total time of flight.
(a) At the peak, \(v = 0\). Use \(v^2 = u^2 - 2gh\): \(0 = 400 - 20h \Rightarrow h = 20\) m.
(b) Time to peak: \(t_{up} = u/g = 2\) s. By symmetry, total flight = 4 s.
Example 5 — Bus with two-phase motion
A bus accelerates uniformly from rest at 1 m/s² for 12 s, then decelerates at 2 m/s² until it stops. Find (a) maximum velocity, (b) total distance covered.
Phase 1: \(v_1 = 0 + 1 \times 12 = 12\) m/s. \(s_1 = \tfrac{1}{2}(1)(12)^2 = 72\) m.
Phase 2: \(u' = 12, v' = 0, a' = -2\). Use \(v'^2 = u'^2 + 2a's_2 \Rightarrow 0 = 144 - 4 s_2 \Rightarrow s_2 = 36\) m.
(a) \(v_{\max} = 12\) m/s. (b) Total \(s = 72 + 36 = 108\) m.
Example 6 — From v–t graph
A v–t graph rises from 0 m/s at t=0 to 30 m/s at t=10 s, stays flat at 30 m/s till t=20 s, then drops to 0 at t=25 s. Find total displacement.
Area = ½(10)(30) + (30)(10) + ½(5)(30) = 150 + 300 + 75 = 525 m.
Example 7 — Distance in nth second
A body starts from rest with acceleration 4 m/s². Find the distance covered in the 5th second.
Distance in nth second \(= u + \tfrac{a}{2}(2n-1) = 0 + 2(2 \times 5 - 1) = 2 \times 9 = 18\) m.
Example 8 — Two trains catching up
Train A moves at constant 20 m/s. Train B, 600 m behind, starts from rest with acceleration 1 m/s² in the same direction. After how long does B catch up?
Distance covered by A in time t = 20t + 600 (B must cover this much from B's start).
Distance by B = \(\tfrac{1}{2}(1)t^2 = 0.5t^2\).
Set equal: \(0.5 t^2 = 20 t + 600 \Rightarrow t^2 - 40 t - 1200 = 0 \Rightarrow t = \dfrac{40 \pm \sqrt{1600+4800}}{2} = \dfrac{40 \pm 80}{2}\).
Taking positive root, \(t = 60\) s.
Example 9 — Circular motion speed
A satellite revolves around Earth in a circular orbit of radius 7000 km, completing one revolution in 1.6 h. Compute its orbital speed (in km/s).
\(v = \dfrac{2\pi r}{T} = \dfrac{2 \times 3.14 \times 7000}{1.6 \times 3600} \approx \dfrac{43960}{5760} \approx 7.63\) km/s.
Example 10 — Athlete on circular track (NCERT-style)
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
\(T = 40\) s; total time \(= 140\) s, so number of rounds \(= 140/40 = 3.5\).
Distance: 3.5 × circumference = 3.5 × π × 200 = 700 π ≈ 2200 m.
Displacement: after 3 full rounds the athlete is back at start. The remaining 0.5 round leaves him diametrically opposite the start; displacement = diameter = 200 m.
Competency-Based Questions
A motorbike starts from rest and accelerates uniformly at 2 m/s² along a straight road. The rider notes the time and the distance covered.
Q1. The velocity attained after 8 s is — L2
(b) 16 m/s. v = u + at = 0 + 2(8) = 16 m/s.
Q2. Distance covered in 8 s using \(s = ut + \tfrac{1}{2}at^2\) is — L3
s = 0 + ½(2)(64) = 64 m.
Q3. To reach 30 m/s from rest, the bike needs — L3
t = v/a = 30/2 = 15 s.
Q4. On a v–t graph for this trip, the slope represents — L2
Acceleration (slope of v–t graph = a = 2 m/s²).
Q5. If the rider then brakes uniformly and stops in 4 s, the braking distance is — L4
From 16 m/s to 0 in 4 s: \(a = -4\) m/s²; \(s = ut + \tfrac{1}{2}at^2 = 16(4) + \tfrac{1}{2}(-4)(16) = 64 - 32 = 32\) m.
Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion: The area under a velocity–time graph gives the displacement of the object.
Reason: Displacement equals velocity multiplied by time.
(A) Each thin strip of width \(dt\) has area \(v\,dt\), summed gives total displacement.
Assertion: An object in uniform circular motion has acceleration even though its speed is constant.
Reason: The direction of velocity changes continuously in circular motion.
(A) Acceleration is rate of change of velocity (a vector); changing direction means changing velocity → acceleration ≠ 0.
Assertion: The slope of a distance–time graph for a stationary body is zero.
Reason: A stationary body has zero speed.
(A) Slope of d–t graph = speed; if speed is zero, the line is horizontal.
Frequently Asked Questions — Equations, Graphs & Circular Motion
What is equations, graphs & circular motion in Class 9 Science (CBSE/NCERT)?
Equations, Graphs & Circular Motion is a key topic in NCERT Class 9 Science Chapter 4 — Describing Motion Around Us. It explains three equations of motion, drawing and interpreting distance-time and velocity-time graphs, and uniform circular motion. Core ideas covered include equations of motion, v = u + at, s = ut + 1/2 at^2, v^2 = u^2 + 2as. Mastering this subtopic is essential for scoring well in the CBSE Class 9 Science exam and for building a strong foundation for the Class 10 board exam, because these concepts repeatedly appear in MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.
Why is equations of motion important in NCERT Class 9 Science?
Equations of motion is important in NCERT Class 9 Science because it forms the foundation for understanding equations, graphs & circular motion in Chapter 4 — Describing Motion Around Us. Without a clear idea of equations of motion, students cannot answer higher-order CBSE questions involving v = u + at, s = ut + 1/2 at^2, v^2 = u^2 + 2as. School and competitive papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link equations of motion to real-life situations. Building clarity here pays off directly in marks at Class 9 and again in the Class 10 board exam.
How is equations, graphs & circular motion tested in the Class 9 Science CBSE exam?
The CBSE Class 9 Science exam tests equations, graphs & circular motion through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or derivations) and 4-mark competency-based questions. Expect direct questions on equations of motion, v = u + at, s = ut + 1/2 at^2 and application-based questions drawn from NCERT activities. Students who follow the NCERT Exploration textbook thoroughly and practise this chapter's questions consistently score in the 90%+ range.
What are the key terms to remember for equations, graphs & circular motion in Class 9 Science?
The key terms to remember for equations, graphs & circular motion in NCERT Class 9 Science Chapter 4 are: equations of motion, v = u + at, s = ut + 1/2 at^2, v^2 = u^2 + 2as, distance-time graph, velocity-time graph. Each of these concepts carries exam weightage and regularly appears in the CBSE Class 9 paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 9 Science exam.
Is Equations, Graphs & Circular Motion included in the Class 9 Science syllabus for 2025–26 CBSE?
Yes, Equations, Graphs & Circular Motion is part of the NCERT Class 9 Science syllabus (2025–26) prescribed by CBSE under the new NCERT Exploration textbook. It falls under Chapter 4 — Describing Motion Around Us — and is examined in the annual paper. The current syllabus retains the full treatment of equations of motion, v = u + at, s = ut + 1/2 at^2 as per the NCERT textbook. Because CBSE bases every Class 9 question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.
How should I prepare equations, graphs & circular motion for the CBSE Class 9 Science exam?
Prepare equations, graphs & circular motion for the CBSE Class 9 Science exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of equations of motion, v = u + at, s = ut + 1/2 at^2. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practise competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the exam.
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