NCERT SOLUTIONS • CHAPTER 4

Given: Distance from home to shop = $250 \text{ m}$

Path followed:

1. Home $\to$ Shop = $250 \text{ m}$
2. Shop $\to$ Home = $250 \text{ m}$
3. Home $\to$ Shop = $250 \text{ m}$
4. Shop $\to$ Home = $250 \text{ m}$

1. Total Distance Travelled:

$$\text{Distance} = 250 \text{ m} + 250 \text{ m} + 250 \text{ m} + 250 \text{ m} = 1000 \text{ m} = 1 \text{ km}$$

2. Displacement from Home:

Since his final position is back at home (same as initial position), the net change in position is zero.

$$\text{Displacement} = 0 \text{ m}$$

Given: Height per floor = $3 \text{ m}$

• Ground floor to 4th floor height = $4 \times 3 \text{ m} = 12 \text{ m}$
• 4th floor to 2nd floor height = $2 \times 3 \text{ m} = 6 \text{ m}$

(i) Total Vertical Distance Travelled:

$$\text{Distance} = 12 \text{ m (upwards)} + 6 \text{ m (downwards)} = 18 \text{ m}$$

(ii) Displacement from Starting Point (Ground Floor):

The student's final position is on the 2nd floor, which is $2 \times 3 \text{ m} = 6 \text{ m}$ above the ground floor.

$$\text{Displacement} = +6 \text{ m (in the upward direction)}$$

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

Explanation:

• Acceleration is defined as the rate of change of velocity ($\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$). Velocity is a vector quantity that has both magnitude (speed) and direction.
• If the girl is riding along a curved path or circular track while keeping the speedometer reading constant, the direction of motion continuously changes at every instant.
• A continuous change in direction results in a continuous change in velocity, producing a non-zero centripetal acceleration directed towards the center of curvature.

Given:

• Initial velocity, $u = 0 \text{ m/s}$
• Final velocity, $v = 24 \text{ m/s}$
• Time interval, $t = 6 \text{ s}$

1. Average Acceleration ($a$):

$$a = \frac{v - u}{t} = \frac{24 - 0}{6} = 4 \text{ m/s}^2$$

2. Distance Travelled ($s$):

Using $s = ut + \frac{1}{2}at^2$:

$$s = (0 \times 6) + \frac{1}{2} \times 4 \times (6)^2 = \frac{1}{2} \times 4 \times 36 = 72 \text{ m}$$

Therefore, the average acceleration is $4\text{ m/s}^2$ and the distance travelled is $72\text{ m}$.

Given:

• Initial velocity, $u = 28 \text{ m/s}$
• Final velocity, $v = 0 \text{ m/s}$ (comes to a stop)
• Distance travelled, $s = 98 \text{ m}$

1. Acceleration ($a$):

Using third equation of motion ($v^2 = u^2 + 2as$):

$$0^2 = (28)^2 + 2 \cdot a \cdot 98$$ $$0 = 784 + 196 a \implies 196 a = -784 \implies a = \frac{-784}{196} = -4 \text{ m/s}^2$$

(The negative sign indicates retardation/deceleration).

2. Time taken ($t$):

Using first equation of motion ($v = u + at$):

$$0 = 28 + (-4)t \implies 4t = 28 \implies t = \frac{28}{4} = 7 \text{ s}$$

Therefore, the acceleration is $-4\text{ m/s}^2$ and the time taken to stop is $7\text{ s}$.

Answer: No, objects A and B never have equal velocity.

Justification:

• In a position-time ($s-t$) graph, the slope of the line ($\frac{\Delta s}{\Delta t}$) represents the velocity of the object.
• In Fig. 4.27, both lines for A and B are straight lines with completely different, non-parallel slopes (slope of line A is steeper than slope of line B).
• Since line A always maintains a steeper slope than line B at all times, object A always moves with a higher constant velocity than object B. They intersect at $t=5\text{ s}$, which represents the instant they cross each other's position, not when their velocities become equal.

Correct Options: (i) and (ii)

Explanation:

• At $t = 0\text{ s}$, both objects start at the origin ($s = 0\text{ m}$). At $t = 10\text{ s}$, both objects reach the same final position.
• Since initial and final positions are identical, their net displacements are equal $\implies \text{Average Velocity } v_{av} = \frac{\text{Displacement}}{10\text{ s}}$ is equal for both. (Option i is correct).
• Since motion is along a straight line in one direction without turning back, total distance equals displacement $\implies \text{Average Speed}$ is equal for both. (Option ii is correct).

Given:

• Initial velocity, $u = 54 \text{ km/h} = 54 \times \frac{5}{18} = 15 \text{ m/s}$
• Final velocity, $v = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}$
• Time, $t = 36 \text{ s}$

1. Acceleration ($a$):

$$a = \frac{v - u}{t} = \frac{10 - 15}{36} = -\frac{5}{36} \text{ m/s}^2$$

2. Distance Travelled ($s$):

$$s = ut + \frac{1}{2}at^2 = (15 \times 36) + \frac{1}{2} \times \left(-\frac{5}{36}\right) \times (36)^2$$ $$s = 540 - \left(\frac{5}{2} \times 36\right) = 540 - 90 = 450 \text{ m}$$

Therefore, the distance travelled while slowing down is 450 m.

The motion consists of three distinct stages:

Stage 1: Uniform Acceleration ($u=0, v=20\text{ m/s}, t_1=5\text{ s}$)

$$s_1 = \left(\frac{u + v}{2}\right) \times t_1 = \left(\frac{0 + 20}{2}\right) \times 5 = 10 \times 5 = 50 \text{ m}$$

Stage 2: Constant Velocity ($v=20\text{ m/s}, t_2=10\text{ s}$)

$$s_2 = v \times t_2 = 20 \text{ m/s} \times 10 \text{ s} = 200 \text{ m}$$

Stage 3: Deceleration to Rest ($u=20\text{ m/s}, v=0, t_3=6\text{ s}$)

$$s_3 = \left(\frac{u + v}{2}\right) \times t_3 = \left(\frac{20 + 0}{2}\right) \times 6 = 10 \times 6 = 60 \text{ m}$$

Total Distance Travelled ($s_{total}$):

$$s_{total} = s_1 + s_2 + s_3 = 50 \text{ m} + 200 \text{ m} + 60 \text{ m} = 310 \text{ m}$$

Therefore, the total distance travelled by the car is 310 m.

Given:

• Initial velocity, $u = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}$
• Reaction time, $t_{react} = 0.5 \text{ s}$
• Deceleration, $a = -2.5 \text{ m/s}^2$
• Final velocity, $v = 0 \text{ m/s}$
• Distance to obstacle = $30 \text{ m}$

1. Distance covered during reaction time ($s_1$):

$$s_1 = u \times t_{react} = 10 \text{ m/s} \times 0.5 \text{ s} = 5 \text{ m}$$

2. Distance covered while braking ($s_2$):

Using $v^2 = u^2 + 2as_2$:

$$0^2 = (10)^2 + 2(-2.5)s_2 \implies 0 = 100 - 5 s_2 \implies 5 s_2 = 100 \implies s_2 = 20 \text{ m}$$

3. Total Stopping Distance ($s_{stop}$):

$$s_{stop} = s_1 + s_2 = 5 \text{ m} + 20 \text{ m} = 25 \text{ m}$$

Conclusion:

Since the total stopping distance ($25\text{ m}$) is less than the distance to the obstacle ($30\text{ m}$), yes, the bus will safely stop 5 m before reaching the obstacle.

Yes, an object kept on the Earth can be considered to be at rest, depending on the chosen Frame of Reference.

Explanation:

• Motion and Rest are relative concepts. An object's state depends entirely on the reference point chosen by the observer.
• With respect to an Earth-bound frame of reference: An object sitting on a table does not change its position relative to the room or building over time. Hence, it is at rest.
• With respect to a Sun-centered (Astronomical) frame of reference: Since the Earth revolves around the Sun at $\approx 30\text{ km/s}$ and rotates on its axis, the object is in motion together with the Earth.

Data extracted from Fig. 4.30:

• Stage 1 ($0\text{ s} \to 20\text{ s}$): Accelerates from $0$ to $3\text{ m/s}$.
• Stage 2 ($20\text{ s} \to 100\text{ s}$): Constant velocity $v = 3\text{ m/s}$ for $80\text{ s}$.
• Stage 3 ($100\text{ s} \to 120\text{ s}$): Velocity decreases from $3\text{ m/s}$ to $2\text{ m/s}$ in $20\text{ s}$.

1. Displacement Calculations:

• (i) Constant Velocity Region ($20\text{ s} \to 100\text{ s}$):
  $\text{Displacement}_2 = \text{Area of Rectangle} = 3 \text{ m/s} \times (100 - 20) \text{ s} = 3 \times 80 = \mathbf{240\text{ m}}$.
• (ii) Decreasing Velocity Region ($100\text{ s} \to 120\text{ s}$):
  $\text{Displacement}_3 = \text{Area of Trapezium} = \frac{3 + 2}{2} \times (120 - 100) = \frac{5}{2} \times 20 = \mathbf{50\text{ m}}$.
• Stage 1 Region ($0\text{ s} \to 20\text{ s}$):
  $\text{Displacement}_1 = \text{Area of Triangle} = \frac{1}{2} \times 20 \times 3 = 30\text{ m}$.

$$\text{Total Displacement in 120 s} = 30 + 240 + 50 = \mathbf{320\text{ m}}$$

2. Average Acceleration over 120 s ($a_{av}$):

$$a_{av} = \frac{v_{final} - u_{initial}}{t_{total}} = \frac{2 \text{ m/s} - 0 \text{ m/s}}{120 \text{ s}} = \frac{2}{120} = \frac{1}{60} \approx \mathbf{0.0167\text{ m/s}^2}$$

Graph Data (Velocity in km/h vs. Time in hours):

• $t = 0 \to 2\text{ h}$: Velocity increases linearly from $6.5\text{ km/h}$ to $7.5\text{ km/h}$.
• $t = 2 \to 4\text{ h}$: Velocity constant at $7.5\text{ km/h}$.
• $t = 4 \to 6\text{ h}$: Velocity decreases linearly from $7.5\text{ km/h}$ to $6.0\text{ km/h}$.

Distance Calculation via Area Under Graph:

• $\text{Area}_1 (0-2\text{ h}) = \frac{6.5 + 7.5}{2} \times 2 = 14 \text{ km}$
• $\text{Area}_2 (2-4\text{ h}) = 7.5 \times 2 = 15 \text{ km}$
• $\text{Area}_3 (4-6\text{ h}) = \frac{7.5 + 6.0}{2} \times 2 = 13.5 \text{ km}$

$$\text{Total Distance} = 14 \text{ km} + 15 \text{ km} + 13.5 \text{ km} = \mathbf{42.5\text{ km}}$$

Given:

• Stage 1: Constant velocity $u = 6 \text{ m/s}$ for $t_1 = 2 \text{ min} = 120 \text{ s}$
• Stage 2: Acceleration $a = 1 \text{ m/s}^2$ for $t_2 = 6 \text{ s}$ (from $t = 120\text{ s}$ to $126\text{ s}$)
• Final velocity at $t = 126\text{ s}$: $v = u + a t_2 = 6 + (1 \times 6) = 12 \text{ m/s}$

Displacement Calculation:

• Displacement in first 120 s ($s_1$):
  $s_1 = 6 \text{ m/s} \times 120 \text{ s} = 720 \text{ m}$
• Displacement in next 6 s ($s_2$):
  $s_2 = ut_2 + \frac{1}{2} a t_2^2 = (6 \times 6) + \frac{1}{2}(1)(6)^2 = 36 + 18 = 54 \text{ m}$

$$\text{Total Displacement} = s_1 + s_2 = 720 \text{ m} + 54 \text{ m} = \mathbf{774\text{ m}}$$

Calculations:

• Car A: $u_A = 0, v_A = 5 \text{ m/s}, t_A = 5 \text{ s} \implies a_A = \frac{5-0}{5} = 1 \text{ m/s}^2$
• Car B: $u_B = 0, v_B = 3 \text{ m/s}, t_B = 10 \text{ s} \implies a_B = \frac{3-0}{10} = 0.3 \text{ m/s}^2$

Displacement Calculations from Graph:

• Displacement of Car A in 5 s ($s_A$):
  $s_A = \text{Area under triangle A} = \frac{1}{2} \times 5 \text{ s} \times 5 \text{ m/s} = \mathbf{12.5\text{ m}}$.
• Displacement of Car B in 10 s ($s_B$):
  $s_B = \text{Area under triangle B} = \frac{1}{2} \times 10 \text{ s} \times 3 \text{ m/s} = \mathbf{15\text{ m}}$.

Given:

• Time duration, $\Delta t = 6:00 \text{ PM to } 7:30 \text{ PM} = 1.5 \text{ hours} = 90 \text{ minutes} = 5400 \text{ s}$
• Radius of minute hand circular path, $r = 7 \text{ cm} = 0.07 \text{ m}$
• Number of complete revolutions in $1.5\text{ hours} = 1.5 \text{ revolutions}$

(i) Distance Travelled ($s$):

$$s = 1.5 \times (2 \pi r) = 1.5 \times 2 \times \frac{22}{7} \times 7 \text{ cm} = 66 \text{ cm} = 0.66 \text{ m}$$

(ii) Displacement ($\vec{d}$):

At 6:00 PM, the minute hand points at '12'. At 7:30 PM (after $1.5$ revolutions), it points at '6'. The net displacement is the straight line diameter between '12' and '6'.

$$\text{Displacement} = 2r = 2 \times 7 \text{ cm} = 14 \text{ cm} = 0.14 \text{ m (pointing downwards towards '6')}$$

(iii) Average Speed ($v_{speed}$):

$$v_{speed} = \frac{\text{Distance}}{\text{Time}} = \frac{66 \text{ cm}}{90 \text{ min}} = 0.733 \text{ cm/min} = \frac{0.66 \text{ m}}{5400 \text{ s}} \approx 1.22 \times 10^{-4} \text{ m/s}$$

(iv) Average Velocity ($v_{velocity}$):

$$v_{velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{14 \text{ cm (downward)}}{90 \text{ min}} \approx 0.156 \text{ cm/min (downward)}$$