Solutions-NCERT-Class-9-Science (Exploration)-Chapter-4-Describing Motion Around Us-CBSE

When Displacement is Zero:

Displacement is the net change in position between two given instants of time. Therefore, the displacement of the athlete is zero whenever the athlete returns to the initial starting position (origin O).

Total Distance Travelled:

If the athlete runs from the origin O to point A (100 m) and then returns all the way back to O (100 m), the total distance travelled is the sum of the distances covered.

Total Distance = OA + AO

= 100 m + 100 m

= 200 m

(i) Total distance travelled

Justification:

• Fuel consumption depends on the engine work required to continuously cover a path length throughout the journey. Total distance travelled is a scalar quantity representing the actual total path length covered.
• Displacement, on the other hand, measures only the net change in position between the start and end points. If a vehicle makes a complete round trip and returns to its starting point, its net displacement is zero, but fuel is still consumed to move the vehicle across the total distance travelled.

Is it straight-line motion?

• Yes, because the ball moves along a single straight path along the inclined track, which is linear motion (motion in a straight line).

Can it be depicted using a horizontal line?

• For motion along a straight path, positions can be marked along a one-dimensional reference line with an origin ‘O’ and directional signs (+) and (-)

Are total distance and magnitude of displacement equal or different at A, B, C, and D?

• They are equal at positions A, B, C, and D. For motion in a straight line, the total distance travelled and the magnitude of displacement are always equal provided the object moves in one direction without turning back.

Given Data:

• Leg 1: Distance = 200 km North, Time = 3 h
• Leg 2: Distance = 200 km South, Time = 2 h

Calculation of Average Speed:

Total Distance Travelled = d₁ + d₂ = 200 km + 200 km = 400 km

Total Time Interval = t₁ + t₂ = 3 h + 2 h = 5 h

Average Speed = \(\frac{\text{Total Distance Travelled}}{\text{Total Time}}\) = \(\frac{400\,km}{5\,h}\) = 80 km h⁻¹

Calculation of Average Velocity:

The vehicle travels 200 km North and then 200 km South. Therefore, it returns to its exact starting point.

Hence, the net displacement = 0 km.

Average Velocity = \(\frac{Displacement}{\text{Total Time}}\) = \(\frac{0\,km}{5\,h}\) = 0 km h⁻¹

(i) Equal Magnitude Condition:

• The magnitude of average velocity is equal to average speed when an object moves in a straight line in one direction only without turning back. Under this condition, total distance travelled equals the magnitude of displacement.

(ii) Zero Velocity with Non-Zero Speed Condition:

• The average velocity magnitude is zero while average speed is non-zero when an object moves over a path (covering a non-zero distance) and returns to its initial starting point or origin. In this situation, the net displacement is zero while the total distance travelled is greater than zero.

Path: Home → Shop (250 m) → Home (250 m) → Shop (250 m) → Home (250 m)

∴ Total distance travelled = 250 + 250 + 250 + 250 = 1000 m

Since he ends up back at home (his starting point), the net change in position is zero.

∴ Displacement from home = 0 m

Ground floor → 4th floor: rises through 4 floors = 4 × 3 m = 12 m (upward)

4th floor → 2nd floor: comes down through 2 floors = 2 × 3 m = 6 m (downward)

(i) Total vertical distance travelled = 12 m + 6 m = 18 m

(ii) Net displacement = position of 2nd floor − position of ground floor

= 2 × 3 m – 0 m = 6 m, directed upward, since the student ends up 2 floors above the start.

Yes, it is possible. A speedometer shows only the magnitude of velocity (speed) — it gives no information about direction.

If the scooter is moving on a curved or circular road (or turning) at a constant speed, its speed does not change, but the DIRECTION of its velocity keeps changing continuously.

This change in direction means her velocity is changing, which requires

acceleration.

Given: u = 0 m/s, v = 24 m/s, t = 6 s

Average acceleration, a = \(\frac{v-u}{t}\) = \(\frac{24-0}{6}\) = 4 m/s²

Distance, s = ut + \(\frac{1}{2}\)at² = 0 × 6 + \(\frac{1}{2}\)(4)(6²) = 0 +  (4)(36) = 72 m

Given: u = 28 m/s, v = 0 m/s, s = 98 m

(i) To find the acceleration

Using v² = u² + 2as:  0 = (28)² + 2 × a × 98

0 = 784 + 196a 

∴ a = \(\frac{-784}{196}\) = −4 m/s²

(ii) To find time taken to come to a stop

Using v = u + at: 

0 = 28 + (−4)t 

∴ t = 28/4 = 7 s

Both position-time graphs of A and B are STRAIGHT LINES, so each object moves with its own constant velocity throughout (velocity = slope of the line).

Since the two lines have different (unequal) slopes, the velocities of A and B are different at every instant — they are never equal.

The point where the two lines cross (around t = 5 s) only means A and B are at the SAME POSITION at that instant — it does not mean their velocities are equal.

Both A and B start at the same position and end at the same position after 10 s, so both have the same displacement over the same time interval.

Average velocity = \(\frac{Displacement}{Time}\)  → since displacement and time are equal for both, their average velocities are equal.

• So (i) is CORRECT.

Since both A and B move steadily forward without reversing direction in this graph, the total distance travelled by each equals the magnitude of its displacement. As both displacements (and hence distances) are equal over the same 10 s, their average speeds are also equal.

• So (ii) is CORRECT.

Options (iii) and (iv) are INCORRECT, because they wrongly assume the total distances covered by A and B are different — in fact, both cover the same net distance in this graph (even though B speeds up/slows down along the way, i.e., B has a curved, non-uniform graph while A is uniform).

Correct options: (i) and (ii).

u = 54 km/h = 54 × \(\frac{1000}{3600}\) = 15 m/s

v = 36 km/h = 36 × \(\frac{1000}{3600}\) = 10 m/s;  t = 36 s

To find distance travelled :

Using s = \(\frac{1}{2}\)(u + v)t = \(\frac{1}{2}\)(15 + 10)(36) =  (25)(36) = 450 m

Phase 1 : (0 to 5 s): u = 0, v = 20 m/s, t = 5 s

s₁ = \(\frac{1}{2}\)(u + v)t = \(\frac{1}{2}\)(0 + 20)(5) = 50 m

Phase 2 : (constant velocity, 10 s): s₂ = v × t = 20 × 10 = 200 m

Phase 3 : (braking, 6 s): u = 20 m/s, v = 0, t = 6 s

s₃ = \(\frac{1}{2}\)(20 + 0)(6) = 60 m

Total distance = s₁ + s₂ + s₃ = 50 + 200 + 60 = 310 m

Speed, u = 36 km/h = 10 m/s

Distance covered during reaction time (bus moves at constant 10 m/s for 0.5 s, brakes not yet applied):

Reaction distance = 10 × 0.5 = 5 m

Remaining distance to obstacle = 30 − 5 = 25 m

After brakes are applied: u = 10 m/s, a = −2.5 m/s², v = 0

Using v² = u² + 2as: 

0 = (10)² + 2(−2.5)s 

⇒  0 = 100 − 5s 

⇒  s = \(\frac{100}{5}\) = 20 m

Braking distance (20 m) is less than the remaining 25 m available.

Yes, the bus will be able to stop before reaching the obstacle — with about 5 m to spare.

Whether an object is 'at rest' or 'in motion' depends entirely on the reference point chosen — rest and motion are relative, not absolute.

If we take a point on the Earth's surface as the reference point, an object lying on the ground does not change its position relative to the Earth — so it can be considered at rest with respect to the Earth.

However, if we take the Sun (or a point in space) as the reference point, the same object is in motion, because the Earth itself revolves around the Sun, carrying every object on its surface along with it.

Conclusion: The same object can be 'at rest' with respect to one reference point and 'in motion' with respect to another — there is no single, absolute answer.

(Readings below are taken from the graph; use your own graph's exact values if they differ slightly.)

In a velocity-time graph, the total displacement is equal to the area under the curve. We can determine this by dividing the area into three simple geometric regions, a triangle (0 to 20 s), a rectangle (20 to 100 s), and a trapezium (100 to 120 s).

Phase 1: (0–20 s), accelerating: 0 → 3 m/s.

Area = \(\frac{1}{2}\)(20)(3) = 30 m

Phase 2: (20 – 100 s), constant velocity 3 m/s — this is region (i):

Area = 3 × (100 − 20) = 3 × 80 = 240 m

Phase 3: (100 – 120 s), decreasing 3 → 2 m/s — this is region (ii):

Area = \(\frac{1}{2}\)(3 + 2)(20) = \(\frac{1}{2}\)(5)(20) = 50 m

Total displacement in 120 s = 30 + 240 + 50 = 320 m

Average acceleration over 120 s = \(\frac{\text{final velocity - initial velocity}}{time}\) = \(\frac{2-0}{120}\)  ≈ 0.017 m/s²

The running distance equals the AREA enclosed between the velocity-time curve and the time axis.

We can approximate the area by breaking it into regular sections

A₁ (0, to 0.6 h, rectangle ) ~ 0.6 x 7 = 4.2 km

A₂ (0.6 to 1.6 h, trapezoid) ~  \(\frac{1}{2}\)(7 + 7.6) × 1.0 = 7.3 km

A3(1.6 to 3.0 h, rectangle) ~1.4 x 7.6 = 10.64 km

A4 (3.0 to 5.5 h, trapezoid ) ~ \(\frac{1}{2}\)(7.6 + 6.4) x 2.5 = 17.5 km

A5 (5.5 to 6.5 h, rectangle) ~1.0x 6.4 = 6.4km

Total estimated distance ~ 4.2 + 73 + 10.64 + 17.5 + 6.4 = 46.04 km

Velocity-time graph: constant at 6 m/s for 120 s, then rising to 12 m/s over the next 6 s.

Phase 1: constant velocity 6 m/s for t = 2 min = 120 s

Displacement, s₁ = v × t = 6 × 120 = 720 m

Phase 2: u = 6 m/s, a = 1 m/s², t = 6 s

Final velocity, v = u + at = 6 + 1(6) = 12 m/s

Displacement, s₂ = ut + \(\frac{1}{2}\)at² = 6(6) + \(\frac{1}{2}\)(1)(6²) = 36 + 18 = 54 m

Total displacement = s₁ + s₂ = 720 + 54 = 774 m

Car A: u = 0, v = 5 m/s, t = 5 s 

⇒  a_A = \(\frac{5-0}{5}\) = 1 m/s²

Car B: u = 0, v = 3 m/s, t = 10 s 

⇒  a_B = \(\frac{3-0}{10}\) = 0.3 m/s²

Velocities of Car A at t = 0, 1, 2, 3, 4, 5 s: 0, 1, 2, 3, 4, 5 m/s

Velocities of Car B at t = 0, 2, 4, 6, 8, 10 s: 0, 0.6, 1.2, 1.8, 2.4, 3.0 m/s

Velocity-time graphs of car A (steeper line, reaches 5 m/s in 5 s) and car B (gentler line, reaches 3 m/s in 10 s).

Displacement = area of the triangle under each line (area = \(\frac{1}{2}\) × base × height):

Car A, over its 5 s: s_A = \(\frac{1}{2}\)(5)(5) = 12.5 m

Car B, over its 10 s: s_B = \(\frac{1}{2}\)(10)(3) = 15 m

Time interval = 6:00 PM to 7:30 PM = 1.5 hours = 90 minutes = 5400 s

The minute's hand completes 1 full revolution every 60 minutes, so in 90 minutes it completes 1.5 revolutions (i.e., it returns to the 12 mark once, then goes exactly halfway round again, ending at the 6 mark).

Radius, R = 7 cm

(i) Distance travelled = 1.5 × circumference = 1.5 × 2πR = 1.5 × 2 × 3.14 × 7 = 1.5 × 43.96 ≈ 65.94 cm

(ii) Displacement: the tip starts at the '12' position and ends at the '6' position — diametrically opposite points on the circle. So displacement = diameter = 2R = 2 × 7 = 14 cm (directed from the 12-mark to the 6-mark).

(iii) Speed = \(\frac{Distance}{time}\) = \(\frac{65.94\,cm}{5400\,s}\) ≈ 0.0122 cm/s

(iv) Velocity (magnitude) = \(\frac{Displacement}{time}\) = \(\frac{14\,cm}{5400\,s}\) ≈ 0.0026 cm/s, directed from the starting point (12) towards the ending point (6).