Problem Types & Solved Examples

6 problem categories. Every type the examiner uses. Broken down step by step. "If this step is wrong, the entire solution fails." — Know exactly where and why.

Direct Formula Conceptual Multi-Step Graph-Based A&R Type

🔬 What is Type 1?

One or two data substituted directly into one formula. Appears in CBSE, NEET. Tests if you know the right formula. Never skip this type — they're free marks if you know your formulas.

Type 1 — Direct Formula

NEET 2019 Style

Easy

Question

A car starts from rest and accelerates uniformly at 4 m/s². Find: (a) velocity after 5 s, (b) distance covered in 5 s.

Given / What Examiner Tests

u = 0 (starts from rest)
a = 4 m/s² (uniform acceleration)
t = 5 s
Tests: Correct equation selection when t is given

Concept Selection

🧠 Which Equation?

Given: u, a, t → Find v: Use v = u + at. Find s: Use s = ut + ½at². No "v²=u²+2as" because t is given and used.

Step-by-Step Solution

1Find final velocity

v = u + at = 0 + (4)(5) = 20 m/s

2Find displacement

s = ut + ½at² = (0)(5) + ½(4)(5²) = 0 + ½(4)(25) = 50 m

Shortcut Insight

🎯 Speed Trick

For u=0: s = ½at². The displacement is proportional to t². Also: v = at, so s = v²/2a. Memorize: starting from rest, after time t, v = at and s = ½at². No manual substitution needed.

Type 1 — Choosing v²=u²+2as

JEE Main 2020 Style

Medium

Question

A bullet of mass 50 g moving at 400 m/s enters a wooden block and comes to rest after penetrating 20 cm. Find the average retardation.

Given

u = 400 m/s, v = 0 (comes to rest)
s = 20 cm = 0.20 m
Find: a (retardation)
Note: Time is NOT given, NOT asked

Concept Selection

🧠 Which Equation?

t is absent → Use v² = u² + 2as immediately. No hesitation needed. This is the standard approach for "penetration", "braking distance" type problems.

Solution

1Apply v² = u² + 2as

0 = (400)² + 2(a)(0.20)

2Solve for a

0 = 160000 + 0.4a → a = −160000/0.4 = −4 × 10⁵ m/s² (retardation)

🎯 Insight

The negative sign means deceleration (opposing motion). Magnitude of retardation = 4 × 10⁵ m/s². Mass was irrelevant here — pure kinematics ignores mass.

🔬 What is Type 2?

No calculation needed. Pure understanding tested. Appears in NEET (1-mark) and JEE assertions. This is where conceptual gaps cost toppers their rank.

Conceptual — Distance vs Displacement

NEET 2018 Style

Easy-Medium

Question

A person walks 10 m east, then 6 m west, then 4 m east. Find (a) total distance, (b) magnitude of displacement from start.

What Examiner Tests

Can you properly assign directions and add vectors in 1D?

Solution

1Set up direction (East = +ve)

Displacements: +10, −6, +4 m

2Total distance (scalar, always +ve)

d = 10 + 6 + 4 = 20 m

3Net displacement (vector)

Δx = +10 − 6 + 4 = +8 m (east)

❌ Mistake Point

Students subtract 6 from distance (getting 14) or forget to add 4. The key: distance = absolute sum of all path lengths. Displacement = algebraic sum with direction.

Conceptual — Speed Increase/Decrease

Medium

Question

A particle moving in the negative x-direction has acceleration in the positive x-direction. What happens to its speed?

Solution

🧠 The Logic

Velocity is negative (−ve direction). Acceleration is positive (+ve direction). They are in OPPOSITE directions → Speed DECREASES. The particle is decelerating despite having positive acceleration. This is the most commonly confused concept.

1Check velocity direction

v < 0 (negative x direction)

2Check acceleration direction

a > 0 (positive x direction)

3v and a are in opposite directions

→ Speed (|v|) decreases. Particle decelerates.

🔬 What is Type 3?

Requires 3+ steps. Different phases of motion. Appears heavily in JEE Main and NEET. This is where most marks are lost — students skip intermediate unknowns.

Multi-Step — Two-Phase Motion

JEE Main 2021 Style

Medium-Hard

Question

A train accelerates uniformly from rest at 2 m/s² for 30 s, then decelerates uniformly at 4 m/s² to rest. Find the total distance covered and total time taken.

What Examiner Tests

Breaking motion into phases. Correctly using final velocity of phase 1 as initial velocity of phase 2.

Solution

1Phase 1: Acceleration phase

u₁=0, a₁=2 m/s², t₁=30s
v₁ = 0 + 2×30 = 60 m/s
s₁ = 0 + ½(2)(30²) = 900 m

2Phase 2: Deceleration phase

u₂ = v₁ = 60 m/s, a₂ = −4 m/s², v₂ = 0
Using v=u+at: 0 = 60 + (−4)t₂ → t₂ = 15 s

3Distance in phase 2

s₂ = (u₂+v₂)/2 × t₂ = (60+0)/2 × 15 = 450 m

4Total distance & time

s_total = 900 + 450 = 1350 m
t_total = 30 + 15 = 45 s

🎯 Key Pattern

In two-phase problems: velocity at end of phase 1 = velocity at start of phase 2. This linking step is the most commonly missed. Without it, you get garbage answers for phase 2.

Multi-Step — Ball Thrown Upward from Building

NEET 2022 Style

Medium-Hard

Question

A ball is thrown upward from the top of a 80 m building with initial velocity 20 m/s. Find: (a) maximum height reached from ground, (b) time to reach ground, (c) velocity when it hits ground. (g = 10 m/s²)

Solution (Up = +ve, origin at rooftop)

1Maximum height from rooftop

v² = u² − 2gh → 0 = (20)² − 2(10)h → h = 20 m above roof
Maximum height from ground = 80 + 20 = 100 m

2Time to hit ground

Taking downward as positive, origin at rooftop: displacement = −80 m (ground is below)
Using up=+ve: s = ut + ½at² → −80 = 20t − 5t²
5t² − 20t − 80 = 0 → t² − 4t − 16 = 0
t = (4 ± √(16+64))/2 = (4 ± √80)/2 = (4 ± 8.94)/2
Taking positive: t = 6.47 s

3Velocity at ground

v = u + at = 20 − 10(6.47) = −44.7 m/s
Speed = 44.7 m/s (downward)

❌ Sign Convention Trap

If you set up = +ve, the final displacement when the ball hits the ground is NEGATIVE (ground is below the starting rooftop). Students often write s = +80 and get wrong time. Displacement is negative here.

🔬 What is Type 4?

Given a graph, extract information. Most tricky type. Tests visual → mathematical translation. "JEE twists kinematics through graphs. Master this, dominate the paper."

Graph-Based — v–t Area Calculation

JEE Main 2023 Style

Hard

Question

A particle moves as shown in the v–t graph below. From t=0 to t=10s: (a) Find displacement. (b) Find total distance. (c) What is acceleration from t=6 to t=10?

Solution

1Identify regions and areas

0→4s: Triangle, base=4, height=8 → A₁ = ½(4)(8) = 16 m (positive)
4→6s: Rectangle, base=2, height=8 → A₂ = 2×8 = 16 m (positive)
6→10s: v goes from 8 to −4 (crosses zero at t=? )
Zero crossing: v = 8 + (−4−8)/(10−6) × (t−6) = 0 → t = 8.67s

2Area 6→8.67s (positive triangle)

A₃ = ½ × 2.67 × 8 = 10.68 m (+ve)

3Area 8.67→10s (negative triangle)

A₄ = ½ × 1.33 × 4 = 2.66 m (−ve)

4Net displacement

s = 16 + 16 + 10.68 − 2.66 = 40.02 m ≈ 40 m

5Total distance

d = 16 + 16 + 10.68 + 2.66 = 45.34 m (all areas positive)

6Acceleration 6→10s

a = slope = Δv/Δt = (−4−8)/(10−6) = −3 m/s²

🎯 Graph Shortcut

When v changes sign, always find the exact zero-crossing time first. Then split the area calculation there. Rushing this step → wrong distance. Take 30 seconds to find crossing time.

🔬 What is Type 5?

Assertion–Reason questions test conceptual depth. Appear in NEET (4 marks), JEE Main. The trick: Assertion and Reason can both be true BUT the Reason may not correctly explain the Assertion.

A&R Answer Code

Option	Assertion	Reason	Does R explain A?
(a)	True	True	YES
(b)	True	True	NO
(c)	True	False	—
(d)	False	True	—

Assertion & Reason — Free Fall

Medium

Question

Assertion (A): A body thrown upward with velocity u has the same speed u when it returns to the point of projection.

Reason (R): The body is in free fall under gravity and the equations of motion are symmetric about the highest point.

Analysis

1Check Assertion

True. By symmetry of free fall (or by v²=u²−2gs; at s=0, v=±u → |v|=u).

2Check Reason

True. Equations of motion with constant g are indeed symmetric about the highest point.

3Does R explain A?

YES. The symmetry IS the reason the return speed equals the launch speed.

✅ Answer: (a)

Both A and R are true, and R is the correct explanation of A.

🔬 What is Type 6?

Case-based questions (CBSE Class 11/12) give a paragraph describing a scenario, then ask 4–5 MCQs from it. You must extract all data from the passage. Most common error: missing a number hidden in a sentence.

Case Study — Car Braking Problem

CBSE 2023 Pattern

Medium

Passage

A driver of a car travelling at 72 km/h sees an obstacle 50 m ahead. The driver reacts and applies brakes after a reaction time of 0.5 s. The car decelerates at 10 m/s² after the brakes are applied. (g = 10 m/s²)

Q1: What is the initial speed in m/s?

1Unit conversion

72 km/h = 72 × (1000/3600) = 20 m/s

Q2: Distance covered during reaction time?

1During reaction time, car moves at constant velocity

s_reaction = 20 × 0.5 = 10 m

Q3: Will the car stop before the obstacle?

1Braking distance after reaction

v²=u²+2as → 0=(20)²+2(−10)s → s = 400/20 = 20 m

2Total stopping distance

10 + 20 = 30 m < 50 m → YES, car stops safely with 20 m to spare.

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