Answer:

The distance to maintain from the truck ahead depends on several factors:

• Your speed: Higher speed requires more distance to stop.
• Braking capacity: How quickly your vehicle can slow down.
• Road conditions: Wet or dry roads affect stopping distance.
• Driver reaction time: Time taken to perceive the danger and apply brakes.
• Vehicle mass: Heavier vehicles need more distance to stop.

The safe distance should be enough to stop your vehicle completely before reaching the truck's position if it suddenly brakes.

Answer:

Yes, the required stopping distance depends significantly on speed. As shown in Example 4.8:

• At 54 km/h (15 m/s), stopping distance = 28.1 m.
• At 108 km/h (30 m/s), stopping distance = 112.5 m.

When speed doubles (from 15 m/s to 30 m/s), the stopping distance becomes four times greater (28.1 m to 112.5 m). This is because stopping distance is proportional to the square of velocity (s ∝ v²). Therefore, higher speed requires maintaining a much larger distance from the vehicle ahead.

Answer:

Displacement becomes zero when the athlete returns to the starting point (reference point O).

Using Fig. 4.4: Starting from O, if the athlete runs from O to A (100 m), then back to B (60 m), and continues back to O (40 m more), the displacement will be zero.

Total distance travelled in that case = OA + AB + BO = 100 m + 60 m + 40 m = 200 m

So, displacement is zero but distance travelled is 200 m.

Answer:

Fuel used depends on (i) Total distance travelled, not displacement.

Justification:

• A vehicle's engine works to cover the actual path length, which is the total distance travelled.
• If a vehicle goes from point A to B and back to A, the displacement is zero but fuel is consumed for the entire trip.
• The engine burns fuel to overcome friction, air resistance, and other forces along the entire path travelled.
• Thus, fuel consumption is proportional to the total distance covered, not the net displacement.

Answer:

Yes, the ball's motion is a straight-line motion because it moves along a straight inclined track.

Yes, the motion can be depicted using a horizontal line as in Fig. 4.3 by representing the inclined path as a straight line.

For a straight line motion without turning back:

• The total distance travelled equals the magnitude of displacement at every position (A, B, C, D).
• Since the ball rolls in one direction only (down the incline), distance and displacement magnitude are equal.
• Distance travelled = |Displacement| = length of path from O to that point.

Answer:

Given:

• First leg: 200 km north in 3 hours.
• Second leg: 200 km south in 2 hours.

Total distance travelled = 200 km + 200 km = 400 km

Total time taken = 3 h + 2 h = 5 h

Displacement = 200 km (north) + 200 km (south) = 0 km

Average speed = Total distance / Total time = 400 km / 5 h = 80 km/h

Average velocity = Displacement / Total time = 0 km / 5 h = 0 km/h

The average speed is 80 km/h, but the average velocity is zero because you returned to the starting point.

Answer:

(i) Magnitude of average velocity equals average speed when:

• The object moves in a straight line in one direction only (without turning back).
• In this case, the total distance travelled equals the magnitude of displacement.
• So, average speed = magnitude of average velocity.

(ii) Magnitude of average velocity is zero while average speed is not zero when:

• The object starts from a point and returns to the same point after some time.
• Example: An athlete running around a circular track and returning to the starting point.
• Displacement = 0, so average velocity = 0, but distance travelled > 0, so average speed > 0.

Answer:

Given: Distance from home to shop = 250 m

Path taken: Home → Shop → Home → Shop → Home

Total distance travelled:

• Home to Shop: 250 m
• Shop to Home: 250 m
• Home to Shop: 250 m
• Shop to Home: 250 m
• Total = 250 + 250 + 250 + 250 = 1000 m

Displacement:

• Starting point = Home
• Ending point = Home
• Displacement = 0 m

Answer:

Given: Height of each floor = 3 m

Path: Ground (0) → 4th floor → 2nd floor

(i) Total vertical distance travelled:

• Ground to 4th floor = 4 floors × 3 m = 12 m
• 4th floor to 2nd floor = (4 − 2) floors × 3 m = 2 × 3 = 6 m
• Total distance = 12 m + 6 m = 18 m

(ii) Displacement from starting point:

• Starting point = Ground floor (0 m)
• Ending point = 2nd floor = 2 × 3 m = 6 m
• Displacement = 6 m − 0 m = 6 m upward

Answer:

Yes, it is possible for the scooter to be accelerating even if the speedometer reading is constant.

Reason:

• Acceleration is the rate of change of velocity, not just speed.
• Velocity has both magnitude (speed) and direction.
• If the speedometer reading is constant but the scooter is turning (changing direction), its velocity is changing even though speed is constant.
• Example: A scooter moving at constant speed around a circular path or on a curved road.
• This is uniform circular motion where speed is constant but velocity changes due to direction change.

Answer:

Given:

• Initial velocity, u = 0 m/s (starts from rest)
• Final velocity, v = 24 m/s
• Time, t = 6 s

Average acceleration:

a = (v − u) / t = (24 − 0) / 6 = 24 / 6 = 4 m/s²

Distance travelled:

Using s = ut + ½at²

s = (0 × 6) + ½ × 4 × (6)²

s = 0 + ½ × 4 × 36 = 2 × 36 = 72 m

Answer: Average acceleration = 4 m/s², Distance travelled = 72 m.

Answer:

Given:

• Initial velocity, u = 28 m/s
• Final velocity, v = 0 m/s (stops)
• Distance, s = 98 m

Acceleration:

Using v² = u² + 2as

0² = 28² + 2 × a × 98

0 = 784 + 196a

196a = −784

a = −784 / 196 = −4 m/s² (negative sign indicates deceleration)

Time taken:

Using v = u + at

0 = 28 + (−4) × t

4t = 28

t = 7 s

Answer:

Yes, objects A and B have equal velocity at the point where their graphs have the same slope.

Justification:

• Velocity is given by the slope of the position-time graph.
• If two graphs have the same slope at a particular time, their velocities are equal.
• In Fig. 4.27, the graphs of A and B intersect at a point where the slopes appear equal.
• At that point, both objects have the same velocity (both speed and direction).

Answer:

Correct options: (i) and (iii)

Explanation:

• (i) Correct: Average velocity = displacement / time. Both A and B start and end at the same positions, so displacement is equal. Thus, average velocity is equal.
• (ii) Incorrect: Average speed = total distance / time. The distance covered may not be equal even if displacement is equal, as one may have taken a longer path.
• (iii) Correct: From the graph, A covers less distance (shorter path length) than B in 10 seconds, so average speed of A is lower.
• (iv) Incorrect: Average speed is total distance/time, not based on segment-wise speeds. The statement is not supported by the graph.

Answer:

Given:

• Initial velocity, u = 54 km/h = 54 × (5/18) = 15 m/s
• Final velocity, v = 36 km/h = 36 × (5/18) = 10 m/s
• Time, t = 36 s

Using the formula:

s = (u + v)/2 × t

s = (15 + 10)/2 × 36

s = (25/2) × 36

s = 12.5 × 36 = 450 m

Answer:

Phase 1: Acceleration (0 to 5 s)

• u = 0, v = 20 m/s, t = 5 s
• s₁ = (u + v)/2 × t = (0 + 20)/2 × 5 = 10 × 5 = 50 m

Phase 2: Constant velocity (5 to 15 s)

• v = 20 m/s, t = 10 s
• s₂ = v × t = 20 × 10 = 200 m

Phase 3: Deceleration (15 to 21 s)

• u = 20 m/s, v = 0, t = 6 s
• s₃ = (u + v)/2 × t = (20 + 0)/2 × 6 = 10 × 6 = 60 m

Total distance:

s = s₁ + s₂ + s₃ = 50 + 200 + 60 = 310 m

Answer:

Given:

• Initial velocity, u = 36 km/h = 36 × (5/18) = 10 m/s
• Distance to obstacle = 30 m
• Reaction time = 0.5 s
• Deceleration = 2.5 m/s²

Distance during reaction time:

s_reaction = u × t = 10 × 0.5 = 5 m

Distance after braking:

Using v² = u² + 2as

0² = 10² + 2 × (−2.5) × s

0 = 100 − 5s

5s = 100

s_braking = 20 m

Total stopping distance:

s_total = s_reaction + s_braking = 5 + 20 = 25 m

Conclusion: Since 25 m < 30 m, the bus will be able to stop before reaching the obstacle.

Answer:

Motion is relative. Whether an object is at rest or in motion depends on the reference point chosen.

If the reference point is the Earth itself:

• An object kept on the Earth is at rest relative to the Earth.
• It does not change its position with respect to nearby objects on Earth.

If the reference point is the Sun:

• The same object is in motion because it moves with the Earth around the Sun.
• The object is also rotating with the Earth on its axis.

Conclusion: An object can be considered at rest if we choose a reference frame attached to the Earth, but it is in motion if we consider a celestial reference frame.

Answer:

To shade areas:

• Constant velocity region: Shade the rectangular area between the graph and time axis where velocity is constant.
• Decreasing velocity region: Shade the area under the sloping portion of the graph (where velocity decreases).

Displacement:

Displacement = Area under velocity-time graph

From the graph (hypothetical values):

• Area of rectangle (constant velocity part) = v × t
• Area of triangle (decreasing velocity part) = ½ × base × height
• Total displacement = sum of all areas

Average acceleration:

a = (v_final − v_initial) / t_total

From the graph: v_initial and v_final at t=0 and t=120 s

Answer:

Distance = Area under velocity-time graph

To estimate the distance:

1. Divide the graph into geometric shapes (rectangles, triangles, trapeziums) based on the curve.
2. Calculate the area of each shape.
3. Sum all areas to get the total distance.

Formula:

• Rectangle area = length × breadth
• Triangle area = ½ × base × height
• Trapezium area = ½ × (sum of parallel sides) × height

The area under the entire curve gives the total distance travelled by the runner.

Answer:

Given:

• Phase 1: Constant velocity, v = 6 m/s, t₁ = 2 min = 120 s
• Phase 2: Acceleration, a = 1 m/s², t₂ = 6 s

Phase 1 displacement:

s₁ = v × t = 6 × 120 = 720 m

Phase 2 displacement:

u = 6 m/s, a = 1 m/s², t = 6 s

s₂ = ut + ½at² = 6 × 6 + ½ × 1 × 6² = 36 + 18 = 54 m

Total displacement:

s = s₁ + s₂ = 720 + 54 = 774 m

Velocity-time graph:

• 0 to 120 s: Horizontal line at v = 6 m/s
• 120 to 126 s: Straight line sloping upward from 6 m/s to 12 m/s (v = 6 + 1×6 = 12 m/s)

Area under graph = total displacement = 774 m

Answer:

Car A:

• u = 0, v = 5 m/s, t = 5 s
• a₁ = (5 − 0)/5 = 1 m/s²

Car B:

• u = 0, v = 3 m/s, t = 10 s
• a₂ = (3 − 0)/10 = 0.3 m/s²

Velocities at different times:

Displacement (area under graph):

Car A (0 to 5 s): Area = ½ × 5 × 5 = 12.5 m

Car B (0 to 10 s): Area = ½ × 10 × 3 = 15 m

Answer:

Given: Time interval = 6:00 PM to 7:30 PM = 90 minutes = 1.5 hours

Length of minute hand, R = 7 cm

(i) Distance travelled:

• Minute hand completes 1.5 full rotations in 90 minutes.
• Distance in one rotation = circumference = 2πR = 2 × π × 7 = 44 cm.
• Total distance = 1.5 × 44 = 66 cm

(ii) Displacement:

• In 90 minutes, the minute hand ends at the 6 o'clock position.
• From 6:00 (6 o'clock) to 7:30 (6 o'clock), it returns to the same position.
• Displacement = 0 cm

(iii) Speed:

• Speed = distance / time = 66 cm / 90 min = 0.733 cm/min

(iv) Velocity:

• Velocity = displacement / time = 0 / 90 = 0 cm/min

Answer:

Observation:

• When spun slowly: both numbers and letters are visible.
• When spun faster: numbers fade or disappear, but letters remain visible.

Reason:

• The linear speed of points on the disc depends on their distance from the centre: v = ω × r
• Numbers at 7 cm from centre have higher linear speed than letters at 4 cm from centre.
• At higher speeds, the eye cannot track the fast-moving numbers, making them appear faded.
• The letters (closer to centre) move slower and can still be seen.

Are speeds different?

• Angular speed (ω) is the same for all points on the disc.
• Linear speed (v) is different because v = ω × r.
• Numbers have greater linear speed than letters.

Answer:

This is an experiential activity.

Expected observations:

• Phone on outstretched palm: The accelerometer will show small fluctuations because the hand is not perfectly steady. Tiny involuntary muscle movements cause small accelerations.
• Phone on the floor: The readings will be much closer to zero because the floor is stable and not moving.

What this tells us:

• In real situations, even when we think we are still, our bodies are in constant motion (breathing, heartbeat, muscle tremors).
• Acceleration can be very small and still be detected by sensitive instruments.
• This has applications in medical research for studying movement disorders like Parkinson's disease.

Answer:

Derivation 1: s = vt − ½ at²

From Eq. (4.4a): v = u + at → u = v − at

Substituting in Eq. (4.4b): s = ut + ½at²

s = (v − at)t + ½at²

s = vt − at² + ½at²

s = vt − ½ at²

Derivation 2: s = ½ (u + v)t

From Eq. (4.4a): t = (v − u)/a

Substituting in Eq. (4.4b): s = u[(v−u)/a] + ½ a [(v−u)/a]²

s = u(v−u)/a + (v−u)²/2a

s = [2u(v−u) + (v−u)²]/2a

s = [2uv − 2u² + v² − 2uv + u²]/2a

s = [v² − u²]/2a

s = [(v−u)(v+u)]/2a

Since t = (v−u)/a, s = ½ (u + v)t

Using trapezium area formula:

In velocity-time graph, displacement = area of trapezium = ½ (sum of parallel sides) × height

= ½ (u + v) × t

s = ½ (u + v)t

Answer:

Effect of scale choice:

• Small scale (expanded axes): The graph appears stretched; variations become more visible.
• Large scale (compressed axes): The graph appears compressed; variations seem smaller.
• Different X and Y scales: Can distort the shape of the graph, making it appear steeper or flatter.

Which scale is better:

• Choose scales that make the graph occupy most of the paper.
• Scales should be convenient for reading values.
• Common scales like 1 cm = 1 s and 1 cm = 5 m are good.
• Using the same scale for both axes is preferred when possible.

Graph plotting apps:

• They automatically adjust axes to fit the data.
• This ensures the graph is readable and shows the trend clearly.