Describing the Motion Around Us: NCERT Solutions & Important Questions

Step 1: Given values
Diameter of track = 200 m, so Radius (\(r\)) = 100 m
Time taken for 1 round = 40 s
Total time given = 2 minutes 20 seconds = \((2 \times 60) + 20\) = 140 s

Step 2: Calculate total rounds completed
Number of rounds = Total time / Time for one round
Number of rounds = \(140 / 40\) = 3.5 rounds

Step 3: Calculate Distance covered
Distance in 1 round (Circumference) = \(2\pi r\) = \(2 \times \frac{22}{7} \times 100\) = 628.57 m
Distance in 3.5 rounds = \(3.5 \times 628.57\) = 2200 m

Step 4: Calculate Displacement
After 3 complete rounds, the athlete is back at the starting point (Displacement = 0).
In the next 0.5 (half) round, the athlete reaches exactly the opposite side of the circular track.
Displacement = Shortest straight line = Diameter of the track = 200 m.

Final Answer: Distance = 2200 m, Displacement = 200 m.

(a) From A to B:
Total distance = 300 m
Total displacement = 300 m (since it's a straight line)
Total time = 2 min 30 s = 150 s
Average Speed = Distance / Time = \(300 / 150\) = 2 m/s
Average Velocity = Displacement / Time = \(300 / 150\) = 2 m/s

(b) From A to C:
Total distance = 300 m (A to B) + 100 m (B to C) = 400 m
Total displacement = 300 m - 100 m = 200 m (A to C)
Total time = 150 s + 60 s = 210 s
Average Speed = Distance / Time = \(400 / 210\) = 1.90 m/s
Average Velocity = Displacement / Time = \(200 / 210\) = 0.95 m/s

Let the distance to the school be \(x\) km.

Time taken to reach school (\(t_1\)) = Distance / Speed = \(\frac{x}{20}\) hours
Time taken to return (\(t_2\)) = Distance / Speed = \(\frac{x}{30}\) hours

Total Distance = \(x + x\) = \(2x\)
Total Time = \(t_1 + t_2\) = \(\frac{x}{20} + \frac{x}{30}\)

Taking LCM (60): Total Time = \(\frac{3x + 2x}{60}\) = \(\frac{5x}{60}\) = \(\frac{x}{12}\) hours

Average Speed = Total Distance / Total Time = \(\frac{2x}{\frac{x}{12}}\) = \(2x \times \frac{12}{x}\) = 24 km/h.

Given:
Initial velocity (\(u\)) = 0 m/s (starts from rest)
Acceleration (\(a\)) = 3.0 m/s²
Time (\(t\)) = 8.0 s
Distance (\(s\)) = ?

Using the Second Equation of Motion:
$$s = ut + \frac{1}{2}at^2$$ $$s = (0 \times 8) + \frac{1}{2} \times 3 \times (8)^2$$ $$s = 0 + \frac{1}{2} \times 3 \times 64$$ $$s = 3 \times 32$$ $$s = 96 \text{ m}$$

The boat travels a distance of 96 m.

Given:
Initial velocity (\(u\)) = 5 m/s
Final velocity at maximum height (\(v\)) = 0 m/s
Acceleration (\(a\)) = -10 m/s² (negative because it opposes the upward motion)
Height (\(s\)) = ?
Time (\(t\)) = ?

To find Height (\(s\)):
Using the Third Equation of Motion:
$$v^2 - u^2 = 2as$$ $$(0)^2 - (5)^2 = 2 \times (-10) \times s$$ $$-25 = -20s$$ $$s = \frac{25}{20} = 1.25 \text{ m}$$

To find Time (\(t\)):
Using the First Equation of Motion:
$$v = u + at$$ $$0 = 5 + (-10)t$$ $$10t = 5$$ $$t = \frac{5}{10} = 0.5 \text{ s}$$

The height attained is 1.25 m, and it takes 0.5 seconds to reach there.

Answer: (d) (Difficulty: Medium)

Answer: (a) - Hint: First convert km/h to m/s by multiplying by \(5/18\) (Difficulty: Medium)

Answer: (c) (Difficulty: Hard)

Answer: (c) Acceleration (Difficulty: Easy)

Answer: (b) Speed - Velocity changes continuously due to a change in direction. (Difficulty: Easy)

Answer: Speed is the distance covered per unit time (scalar quantity, always positive). Velocity is the displacement per unit time (vector quantity, can be positive, negative, or zero).

Answer: Yes. In uniform circular motion, an object moves with constant speed but its direction changes continuously. Since velocity is a vector, a change in direction means a change in velocity, hence it is accelerated.

Answer: The odometer measures the total distance traveled by the automobile. (Difficulty: Easy)

Answer: The magnitude of average velocity equals average speed only when the object moves along a straight line continuously in one direction without turning back.

Answer: (Students must practice drawing the Velocity-Time graph for uniform acceleration).

The distance traveled (\(s\)) is given by the area enclosed by the velocity-time graph.
Area = Area of rectangle + Area of triangle
$$s = (u \times t) + \frac{1}{2} \times (v-u) \times t$$
Since \(v-u = at\) (from first equation), substitute this into the equation:
$$s = ut + \frac{1}{2} \times (at) \times t$$
$$s = ut + \frac{1}{2}at^2$$ (Difficulty: Hard)

Answer:
\(u = 0\)
\(v = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}\)
\(t = 5 \text{ min} = 300 \text{ s}\)

(i) Acceleration: \(a = \frac{v-u}{t} = \frac{20-0}{300} = \frac{1}{15} \text{ m/s}^2\)

(ii) Distance (\(s\)): Using \(v^2 - u^2 = 2as \rightarrow (20)^2 - 0 = 2 \times (\frac{1}{15}) \times s \rightarrow 400 = \frac{2}{15}s \rightarrow s = \frac{400 \times 15}{2} = 3000 \text{ m} = 3 \text{ km}\) (Difficulty: Hard)

a) What is the initial velocity of the ball? (Ans: 0 m/s because it is dropped)

b) What will be its final velocity just before touching the ground?
(Ans: Using \(v^2 - u^2 = 2gs \rightarrow v^2 - 0 = 2 \times 10 \times 20 \rightarrow v^2 = 400 \rightarrow v = 20 \text{ m/s}\))

c) How much time will it take to reach the ground?
(Ans: Using \(v = u + gt \rightarrow 20 = 0 + 10t \rightarrow t = 2 \text{ s}\)) (Difficulty: Medium)

Answer: (A) Both are true, and R correctly explains A (e.g., returning to the starting point). (Difficulty: Easy)

Answer: (C) A is true, but R is false. Acceleration is a vector quantity. (Difficulty: Medium)