NCERT Exercise - Q2
Textbook Exercise Problem
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Step-by-Step Verified Solution:
- Step 1: Side of square field = 10 m. Perimeter = 4 × 10 = 40 m.
- Step 2: Time for 1 complete round = 40 seconds.
- Step 3: Total time given = 2 min 20 s = 2 × 60 + 20 = 140 seconds.
- Step 4: Number of rounds = 140 / 40 = 3.5 rounds.
- Step 5: After 3.5 rounds, the farmer is at the opposite diagonal vertex of the square.
- Step 6: Displacement = Diagonal of square = √(10² + 10²) = √(200) = 10√2 ≈ 14.14 m.
Final Answer: Displacement = 10√2 m (≈ 14.14 m)
Key Concept / Exam Strategy: Displacement is the straight-line distance connecting initial and final coordinates regardless of total path traversed.
Generated Exercise #1
Foundation
Question: A car accelerates uniformly from 18 km/h to 72 km/h in 5 seconds. Calculate: (a) Acceleration, (b) Distance covered during this time.
Core Concept Tested: Application of 1st and 2nd kinematic equations with km/h to m/s conversion
Mentor Hint: 18 km/h = 5 m/s, 72 km/h = 20 m/s.
Step-by-Step Solution:
- Step 1: Convert velocities: u = 18 × (5/18) = 5 m/s; v = 72 × (5/18) = 20 m/s.
- Step 2: a = (v - u) / t = (20 - 5) / 5 = 15 / 5 = 3 m/s².
- Step 3: s = ut + 0.5at² = 5(5) + 0.5(3)(25) = 25 + 37.5 = 62.5 m.
Verified Answer: (a) a = 3 m/s², (b) Distance = 62.5 m
Generated Exercise #2
Intermediate
Question: A stone is thrown vertically upward with a velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached and total time of flight.
Core Concept Tested: Free fall and vertical motion under gravity
Mentor Hint: At highest point, final velocity v = 0, a = -g.
Step-by-Step Solution:
- Step 1: At maximum height, v = 0. Using v² = u² - 2gh: 0 = 40² - 2(10)h => 20h = 1600 => h = 80 m.
- Step 2: Time to ascend: v = u - gt => 0 = 40 - 10t => t = 4 s.
- Step 3: Total time of flight = 2 × 4 = 8 s.
Verified Answer: Maximum Height = 80 m, Total Time of Flight = 8 s
Generated Exercise #3
Intermediate
Question: What does the area under a Velocity-Time (v-t) graph represent? What does its slope represent?
Core Concept Tested: Graphical kinematics calculus intuition
Mentor Hint: Area is integral of v·dt (displacement); slope is dv/dt (acceleration).
Step-by-Step Solution:
- Step 1: Slope of v-t graph = Δv / Δt = Acceleration.
- Step 2: Area under v-t graph = Velocity × Time = Displacement (or distance travelled).
Verified Answer: Area represents Displacement; Slope represents Acceleration.
Generated Exercise #4
JEE-NEET Foundation
Question: A particle covers the first half of a total distance with speed v1 and the second half with speed v2. Prove that the average speed is given by the harmonic mean: v_avg = 2v1v2 / (v1 + v2).
Core Concept Tested: Derivation of harmonic mean average speed for equal distance segments
Mentor Hint: Let total distance = 2d. Total time = d/v1 + d/v2.
Step-by-Step Solution:
- Step 1: Total distance = d + d = 2d.
- Step 2: Time for 1st half t1 = d / v1. Time for 2nd half t2 = d / v2.
- Step 3: Total time T = d(1/v1 + 1/v2) = d(v1 + v2) / (v1v2).
- Step 4: Average speed = Total Distance / Total Time = 2d / [d(v1 + v2) / (v1v2)] = 2v1v2 / (v1 + v2).
Verified Answer: v_avg = 2v1v2 / (v1 + v2)
Generated Exercise #5
Advanced HOTS
Question: A train 100 m long travelling at 72 km/h completely crosses a bridge 1 km long. How much time does it take to cross the bridge?
Core Concept Tested: Relative bridge traversal length addition
Mentor Hint: Total distance to cross = Length of train + Length of bridge.
Step-by-Step Solution:
- Step 1: Train speed = 72 × (5/18) = 20 m/s.
- Step 2: Total distance to clear = 100 m + 1000 m = 1100 m.
- Step 3: Time = Distance / Speed = 1100 / 20 = 55 seconds.
Verified Answer: 55 seconds