Q: Q13. A trolley starts from rest on an inclined plane with a constant acceleration of 2 cm s⁻². At 𝑡 = 0, a ball is dropped from the top of the trolley. After 4 s, the ball reaches the

Let's analyze the motion of the ball.\n\nThe trolley starts from rest at t = 0. A ball is dropped from the top of the trolley *at t = 0*. This implies that at the moment the ball is dropped, its initial velocity is the same as the trolley's initial velocity, which is zero (since the trolley starts from rest).\n\nWhen the ball is 'dropped', it means its initial velocity relative to the ground is zero. Once dropped, the ball is no longer influenced by the trolley's acceleration on the inclined plane. It undergoes free fall under gravity.\n\nThe question states that the ball reaches the 'floor' after 4 s. Assuming 'floor' refers to the ground (or the surface from which it was dropped, which is the starting point of the trolley), the ball's motion is simply vertical free fall.\n\nGiven:\nInitial velocity of the ball (u) = 0 m/s (since it's dropped from rest)\nTime (t) = 4 s\nAcceleration due to gravity (g) = 9.8 m/s²\n\nUsing the equation of motion for displacement: s = ut + (1/2)gt²\n\nDistance covered by the ball (s) = (0 m/s × 4 s) + (1/2 × 9.8 m/s² × (4 s)²)\ns = 0 + (1/2 × 9.8 × 16) m\ns = 0 + (4.9 × 16) m\ns = 78.4 m\n\nThe information about the trolley's acceleration (2 cm s⁻²) and the inclined plane is extraneous to finding the distance of the ball from the starting point of the trolley, as the ball is dropped from rest at t=0 and falls vertically under gravity.

Question 1

Q1. In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a mon

Accepted Answer

A body can be considered approximately a point object if its size is much smaller than the distance it travels or the dimensions of the space in which its motion is being studied. In other words, its internal structure or rotation does not significantly affect the description of its motion.

Let's analyze each option:

(a) **A railway carriage moving without jerks between two stations:** The distance between two stations is typically very large (several kilometers) compared to the size of a railway carriage (tens of meters). In this context, the size and internal structure of the carriage are negligible, and its motion can be described by treating it as a point object.

(b) **A monkey sitting on top of a moving train:** Here, the monkey's size is much smaller than the distance the train travels. Even if we consider the train as a point object, the monkey on top of it can also be considered a point object relative to the large distance covered by the train.

(c) **A spinning cricket ball that turns sharply on hitting the ground:** The spinning motion and the sharp turn (due to interaction with the ground) are highly dependent on the size and rotational motion of the ball. To understand and describe this motion accurately, the ball cannot be treated as a point object. Its dimensions and rotational dynamics are crucial.

(d) **A tumbling beaker that has slipped off the edge of a table:** Similar to the cricket ball, the tumbling motion involves rotation and changes in orientation. The size and shape of the beaker are essential for describing its fall and tumble. Therefore, it cannot be considered a point object.

**Conclusion:** The bodies can be considered approximately a point object in examples (a) and (b).

Question 2

Q2. The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct statements abou

Accepted Answer

Let's analyze the given position-time (x-t) graph (Fig. 3.19) to evaluate each statement.

**Description of Fig. 3.19:**
*   The x-axis represents time (t), and the y-axis represents position (x).
*   The origin (O) represents the school (x=0).
*   Graph A starts at t=0, x=0 and goes to a final position P at time t_A.
*   Graph B starts at t_B (where t_B > 0), x=0 and goes to a final position Q at time t_B_final.
*   The final position P (for A) is at a smaller x-value than the final position Q (for B).
*   The slope of graph A is steeper than the slope of graph B.
*   The lines A and B do not intersect after starting from the origin.

Now let's evaluate each statement:

**(a) A lives closer to the school than B.**
*   From the graph, the final position of A (home P) is at a smaller x-coordinate than the final position of B (home Q). Since x represents the distance from the school (origin O), A's home is closer to the school than B's home.
*   **This statement is correct.**

**(b) A starts from the school earlier than B.**
*   A's graph starts at t = 0 (the y-axis). This means A starts at time 0.
*   B's graph starts at a time t > 0 (it begins to the right of the y-axis). This means B starts later than A.
*   **This statement is correct.**

**(c) A walks faster than B.**
*   In a position-time graph, the slope (gradient) represents the velocity (speed in this case, as motion is in one direction). A steeper slope indicates a higher speed.
*   The slope of graph A is clearly steeper than the slope of graph B.
*   Therefore, A walks faster than B.
*   **This statement is correct.**

**(d) A and B reach home at the same time.**
*   A reaches home at time t_A (the time when A's graph ends at P).
*   B reaches home at time t_B_final (the time when B's graph ends at Q).
*   From the graph, t_A is clearly less than t_B_final. A reaches home much earlier than B.
*   **This statement is incorrect.**

**(e) A overtakes B on the road.**
*   Overtaking means that at some point in time, A and B are at the same position (x-coordinate).
*   On an x-t graph, this would be represented by the two graphs intersecting.
*   Looking at the graph, the lines A and B do not intersect after leaving the school (x=0). B starts later and moves slower, so A is always ahead of B (or A has already reached home before B starts or catches up).
*   **This statement is incorrect.**

**Final Answer:** The correct statements are (a), (b), and (c).

Question 3

Q3. A woman starts from her home at 9.00 am, walks with a speed of 5 km h⁻¹ on a straight road up to her office 2.5 km away, stays at the office till 5.00 pm, and returns home by an au

Accepted Answer

To plot the position-time graph, we need to determine the woman's position at different times during her journey. We will take home as the origin (x=0) and 9:00 am as t=0.

**Given Data:**
*   Starting time from home: 9:00 am (t=0)
*   Walking speed to office (v₁): 5 km h⁻¹
*   Distance to office (d): 2.5 km
*   Stay at office until: 5:00 pm
*   Return speed by auto (v₂): 25 km h⁻¹

**Calculations:**

**Phase 1: Home to Office (Walking)**
*   Distance (d) = 2.5 km
*   Speed (v₁) = 5 km h⁻¹
*   Time taken (t₁) = Distance / Speed = 2.5 km / 5 km h⁻¹ = 0.5 hours
*   0.5 hours = 0.5 * 60 minutes = 30 minutes
*   Arrival time at office: 9:00 am + 30 minutes = 9:30 am
*   Position at 9:30 am: x = 2.5 km

**Phase 2: Stay at Office**
*   Duration: From 9:30 am to 5:00 pm
*   Time spent at office = 5:00 pm - 9:30 am = 7 hours 30 minutes = 7.5 hours
*   During this phase, her position remains constant at x = 2.5 km.

**Phase 3: Office to Home (Auto)**
*   Departure time from office: 5:00 pm
*   Distance (d) = 2.5 km
*   Speed (v₂) = 25 km h⁻¹
*   Time taken (t₂) = Distance / Speed = 2.5 km / 25 km h⁻¹ = 0.1 hours
*   0.1 hours = 0.1 * 60 minutes = 6 minutes
*   Arrival time at home: 5:00 pm + 6 minutes = 5:06 pm
*   Position at 5:06 pm: x = 0 km

**Summary of (time, position) points for plotting:**
1.  **Start from home:** (9:00 am, 0 km)
2.  **Arrive at office:** (9:30 am, 2.5 km)
3.  **Leave office:** (5:00 pm, 2.5 km)
4.  **Arrive at home:** (5:06 pm, 0 km)

**Plotting the x-t Graph:**
*   **X-axis (Time):** We can use hours, with 9:00 am as t=0. So, 9:30 am is t=0.5 h, 5:00 pm is t=8 h, and 5:06 pm is t=8.1 h.
*   **Y-axis (Position):** Position in km.

1.  **Segment 1 (Home to Office):** A straight line from (0 h, 0 km) to (0.5 h, 2.5 km). The slope is positive, representing positive velocity (5 km/h).
2.  **Segment 2 (At Office):** A horizontal line from (0.5 h, 2.5 km) to (8 h, 2.5 km). The slope is zero, representing zero velocity.
3.  **Segment 3 (Office to Home):** A straight line from (8 h, 2.5 km) to (8.1 h, 0 km). The slope is negative, representing negative velocity (returning home, -25 km/h).

**Description of the x-t Graph:**
*   The graph starts at the origin (0,0) representing home at 9:00 am.
*   It rises linearly to a position of 2.5 km at 9:30 am, indicating constant positive velocity (walking to office).
*   From 9:30 am to 5:00 pm, the graph is a horizontal line at x = 2.5 km, indicating that the woman is stationary at the office.
*   From 5:00 pm, the graph falls linearly back to x = 0 km (home) at 5:06 pm, indicating constant negative velocity (returning by auto). The slope of this segment will be much steeper than the first segment, reflecting the higher speed of the auto.

Question 4

Q4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and takes 1

Accepted Answer

The drunkard takes 37 seconds to fall into the pit 13 m away from the start.

Question 5

Q5. A jet aeroplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet aeroplane. What is the speed of the latter with

Accepted Answer

The speed of the products of combustion with respect to an observer on the ground is 1000 km/h in the direction opposite to the jet aeroplane's motion.

Question 6

Q6. A car moving along a straight highway with speed 126 km/h is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does i

Accepted Answer

The retardation of the car is 3.0625 m/s², and it takes 11.43 seconds for the car to stop.

Question 7

Q7. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overt

Accepted Answer

The original distance between the driver of A and the guard of B was 1250 m.

Question 8

Q8. An auto driver moving at a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. Calculate the av

Accepted Answer

The average retarding force on the vehicle is 1162.5 N.

Question 9

Q9. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overt

Accepted Answer

The original distance between the driver of A and the guard of B was 1250 m.

Question 10

Q10. Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km/h in the direction A to B notic

Accepted Answer

The period T of the bus service is 9 minutes, and the speed of the buses is 40 km/h.

Question 11

Q11. A player throws a ball upwards with an initial speed of 29.4 m/s.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and ac

Accepted Answer

Here are the answers to the various parts of the question:

**(a) Direction of acceleration during upward motion:**
During the entire flight (both upward and downward motion), the only acceleration acting on the ball (neglecting air resistance) is the acceleration due to gravity (g). This acceleration always acts vertically downwards, towards the center of the Earth. So, during the upward motion, the direction of acceleration is vertically downwards.

**(b) Velocity and acceleration at the highest point:**
At the highest point of its motion, the ball momentarily stops before changing its direction and starting to fall downwards. Therefore, the velocity of the ball at the highest point is 0 m/s.
However, the acceleration due to gravity still acts on the ball at the highest point. Its magnitude is g = 9.8 m/s², and its direction is vertically downwards.

**(c) Signs of position, velocity, and acceleration based on given conventions:**
Given: x = 0 m and t = 0 s at the highest point. Vertically downward direction is the positive x-axis.

*   **During upward motion (before reaching the highest point):**
    *   **Position (x):** Since the highest point is x=0 and downward is positive, any point below the highest point (during upward or downward motion) will have a positive x-coordinate. As the ball moves upwards towards x=0, its position x is positive (e.g., if it's 5m below the peak, x = +5m). So, position is **positive (+)**.
    *   **Velocity (v):** The ball is moving upwards. Since the positive x-axis is downwards, an upward velocity is in the negative x-direction. So, velocity is **negative (-)**.
    *   **Acceleration (a):** Acceleration due to gravity is always downwards. Since downward is the positive x-direction, acceleration is **positive (+)**.

*   **During downward motion (after reaching the highest point):**
    *   **Position (x):** The ball is moving downwards from x=0. Any point below x=0 (the highest point) will have a positive x-coordinate. So, position is **positive (+)**.
    *   **Velocity (v):** The ball is moving downwards. Since downward is the positive x-direction, velocity is **positive (+)**.
    *   **Acceleration (a):** Acceleration due to gravity is always downwards. Since downward is the positive x-direction, acceleration is **positive (+)**.

**(d) Height the ball rises and time to return to player’s hands:**
Given initial speed, u = 29.4 m/s (upwards).
Acceleration due to gravity, g = 9.8 m/s² (downwards).

**To find the maximum height (H):**
At the highest point, the final velocity v = 0 m/s.
Using the kinematic equation: v² = u² + 2as
Here, a = -g (since upward is positive, and g acts downwards).
0² = (29.4)² + 2 * (-9.8) * H
0 = 864.36 - 19.6 H
19.6 H = 864.36
H = 864.36 / 19.6
H = 44.1 m.

**To find the time to return to the player's hands:**
This is the total time of flight. The time taken to reach the highest point is equal to the time taken to fall back to the initial position.

Time to reach highest point (t_up): Using v = u + at
0 = 29.4 + (-9.8) * t_up
9.8 t_up = 29.4
t_up = 29.4 / 9.8 = 3 s.

Total time of flight (t_total) = 2 * t_up = 2 * 3 s = 6 s.

Alternatively, using s = ut + (1/2)at² for the entire flight:
Displacement s = 0 (since the ball returns to the player's hands).
0 = 29.4 * t_total + (1/2) * (-9.8) * t_total²
0 = t_total * (29.4 - 4.9 * t_total)
This gives two solutions: t_total = 0 (initial point) or 29.4 - 4.9 * t_total = 0.
4.9 * t_total = 29.4
t_total = 29.4 / 4.9 = 6 s.

So, the ball rises to a height of 44.1 m and returns to the player's hands after 6 seconds.

Question 12

Q12. A ball is dropped from a height of 90 m on a floor. At each rebound, the ball loses one-tenth of its speed. Plot the speed-time graph of the ball from t = 0 to 12 s.

Accepted Answer

To plot the speed-time graph, we need to calculate the speed of the ball just before and after each rebound, and the time taken for each fall and rise. The motion is under constant acceleration due to gravity (g = 9.8 m/s²).

**1. First Fall (from 90 m):**
Initial speed (u) = 0 m/s
Height (h) = 90 m
Speed just before impact (v₁) = √(2gh) = √(2 × 9.8 × 90) = √1764 = 42 m/s
Time taken for first fall (t₁) = v₁/g = 42 / 9.8 = 4.286 s

**2. First Rebound:**
Speed after first rebound (v'₁) = v₁ - (1/10)v₁ = (9/10)v₁ = (9/10) × 42 = 37.8 m/s

**3. First Rise (after first rebound):**
Initial speed (u') = 37.8 m/s
Final speed (v') = 0 m/s (at maximum height)
Time taken to reach max height (t'₁) = u'/g = 37.8 / 9.8 = 3.857 s
Maximum height reached (h') = (u')² / (2g) = (37.8)² / (2 × 9.8) = 1428.84 / 19.6 = 72.9 m

**4. Second Fall (from 72.9 m):**
Initial speed (u) = 0 m/s
Speed just before second impact (v₂) = √(2gh') = √(2 × 9.8 × 72.9) = √1428.84 = 37.8 m/s
Time taken for second fall (t₂) = v₂/g = 37.8 / 9.8 = 3.857 s

**Total time elapsed:**
After first impact: t₁ = 4.286 s
After first rise: t₁ + t'₁ = 4.286 + 3.857 = 8.143 s
After second impact: t₁ + t'₁ + t₂ = 4.286 + 3.857 + 3.857 = 12.000 s

Since the total time is exactly 12.000 s, we need to plot the graph up to the second impact.

**Description of the Speed-Time Graph:**
*   **t = 0 to 4.286 s (First Fall):** The speed increases linearly from 0 to 42 m/s. (A straight line segment from (0,0) to (4.286, 42)). The slope is g = 9.8 m/s².
*   **At t = 4.286 s (First Rebound):** The speed instantaneously drops from 42 m/s to 37.8 m/s. (A vertical dashed line or a point discontinuity at t=4.286s, from speed 42 to 37.8).
*   **t = 4.286 s to 8.143 s (First Rise):** The speed decreases linearly from 37.8 m/s to 0 m/s. (A straight line segment from (4.286, 37.8) to (8.143, 0)). The slope is -g = -9.8 m/s².
*   **t = 8.143 s to 12.000 s (Second Fall):** The speed increases linearly from 0 m/s to 37.8 m/s. (A straight line segment from (8.143, 0) to (12.000, 37.8)). The slope is g = 9.8 m/s².

The graph will consist of piecewise linear segments, with instantaneous drops in speed at the moments of impact (rebound). Since it's a speed-time graph, all speed values are positive.

Question 13

Q13. A trolley starts from rest on an inclined plane with a constant acceleration of 2 cm s⁻². At 𝑡 = 0, a ball is dropped from the top of the trolley. After 4 s, the ball reaches the

Accepted Answer

Let's analyze the motion of the ball.

The trolley starts from rest at t = 0. A ball is dropped from the top of the trolley *at t = 0*. This implies that at the moment the ball is dropped, its initial velocity is the same as the trolley's initial velocity, which is zero (since the trolley starts from rest).

When the ball is 'dropped', it means its initial velocity relative to the ground is zero. Once dropped, the ball is no longer influenced by the trolley's acceleration on the inclined plane. It undergoes free fall under gravity.

The question states that the ball reaches the 'floor' after 4 s. Assuming 'floor' refers to the ground (or the surface from which it was dropped, which is the starting point of the trolley), the ball's motion is simply vertical free fall.

Given:
Initial velocity of the ball (u) = 0 m/s (since it's dropped from rest)
Time (t) = 4 s
Acceleration due to gravity (g) = 9.8 m/s²

Using the equation of motion for displacement: s = ut + (1/2)gt²

Distance covered by the ball (s) = (0 m/s × 4 s) + (1/2 × 9.8 m/s² × (4 s)²)
s = 0 + (1/2 × 9.8 × 16) m
s = 0 + (4.9 × 16) m
s = 78.4 m

The information about the trolley's acceleration (2 cm s⁻²) and the inclined plane is extraneous to finding the distance of the ball from the starting point of the trolley, as the ball is dropped from rest at t=0 and falls vertically under gravity.

Question 14

Q14. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h⁻¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5

Accepted Answer

Let's denote the home as the origin (x=0) and the market as x = +2.5 km.

**Part (a) and (b) calculations require breaking down the journey:**

**1. Time to reach market (t₁):**
Distance = 2.5 km, Speed = 5 km/h
t₁ = Distance / Speed = 2.5 km / 5 km/h = 0.5 h = 30 min

**2. Time to return home (t₂):**
Distance = 2.5 km, Speed = 7.5 km/h
t₂ = Distance / Speed = 2.5 km / 7.5 km/h = 1/3 h = 20 min

**Total time for round trip = t₁ + t₂ = 30 min + 20 min = 50 min**

**Total distance for round trip = 2.5 km + 2.5 km = 5 km**

**Total displacement for round trip = 0 km (since he returns to the starting point)**

---

**(i) Time interval: 0 to 30 min**

*   **Position at t = 30 min:** The man reaches the market, 2.5 km from home.
*   **Displacement (Δx):** 2.5 km (from home to market)
*   **Distance (d):** 2.5 km
*   **Time (Δt):** 30 min = 0.5 h

**(a) Magnitude of Average Velocity:**
Average Velocity = |Δx / Δt| = |2.5 km / 0.5 h| = 5 km/h

**(b) Average Speed:**
Average Speed = d / Δt = 2.5 km / 0.5 h = 5 km/h

---

**(ii) Time interval: 0 to 50 min**

*   **Position at t = 50 min:** The man returns home (origin).
*   **Displacement (Δx):** 0 km (final position = initial position)
*   **Distance (d):** 5 km (2.5 km to market + 2.5 km back home)
*   **Time (Δt):** 50 min = 50/60 h = 5/6 h

**(a) Magnitude of Average Velocity:**
Average Velocity = |Δx / Δt| = |0 km / (5/6 h)| = 0 km/h

**(b) Average Speed:**
Average Speed = d / Δt = 5 km / (5/6 h) = 6 km/h

---

**(iii) Time interval: 0 to 40 min**

*   **Position at t = 40 min:**
    *   First 30 min: Man travels to market (2.5 km).     *   Next 10 min (from 30 min to 40 min): Man is walking back home from the market.     *   Distance covered while returning in 10 min: Speed × Time = 7.5 km/h × (10/60 h) = 7.5 km/h × (1/6 h) = 1.25 km.     *   Final position from home = 2.5 km (market) - 1.25 km (returned) = 1.25 km from home.

*   **Displacement (Δx):** 1.25 km (from home)
*   **Distance (d):** 2.5 km (to market) + 1.25 km (back from market) = 3.75 km
*   **Time (Δt):** 40 min = 40/60 h = 2/3 h

**(a) Magnitude of Average Velocity:**
Average Velocity = |Δx / Δt| = |1.25 km / (2/3 h)| = 1.25 × 3 / 2 km/h = 3.75 / 2 km/h = 1.875 km/h

**(b) Average Speed:**
Average Speed = d / Δt = 3.75 km / (2/3 h) = 3.75 × 3 / 2 km/h = 11.25 / 2 km/h = 5.625 km/h

Question 15

Q15. A police van moving on a highway with a speed of 30 km h⁻¹ fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h⁻¹. If the muzzle speed of th

Accepted Answer

To determine the speed with which the bullet hits the thief's car, we need to find the relative speed of the bullet with respect to the thief's car. This is the speed at which the bullet approaches and impacts the car.

**1. Convert all speeds to a consistent unit (m/s):**

*   **Speed of police van (v_p):**
    v_p = 30 km/h = 30 × (1000 m / 3600 s) = 30 × (5/18) m/s = 8.33 m/s (approx 25/3 m/s)

*   **Speed of thief's car (v_t):**
    v_t = 192 km/h = 192 × (1000 m / 3600 s) = 192 × (5/18) m/s = 960/18 m/s = 160/3 m/s = 53.33 m/s

*   **Muzzle speed of bullet (v_b_muzzle):**
    v_b_muzzle = 150 m/s (This is the speed of the bullet relative to the police van).

**2. Calculate the absolute speed of the bullet (v_b) with respect to the ground:**
Since the bullet is fired from the police van in the same direction as the van's motion, the bullet's speed relative to the ground is the sum of the van's speed and the muzzle speed.
v_b = v_p + v_b_muzzle
v_b = (25/3 m/s) + 150 m/s = (25 + 450)/3 m/s = 475/3 m/s = 158.33 m/s

**3. Calculate the relative speed of the bullet with respect to the thief's car (v_b_relative_t):**
Both the bullet and the thief's car are moving in the same direction. The speed at which the bullet hits the car is the difference between the bullet's absolute speed and the car's absolute speed.
v_b_relative_t = v_b - v_t
v_b_relative_t = (475/3 m/s) - (160/3 m/s)
v_b_relative_t = (475 - 160)/3 m/s
v_b_relative_t = 315/3 m/s
v_b_relative_t = 105 m/s

The bullet hits the thief's car with a speed of 105 m/s.

Question 16

Q16. An object moving with a speed of 6.25 m s⁻¹ is decelerated at a rate given by dv/dt = -2.5√v where v is the instantaneous speed. The time taken by the object, to come to rest, woul

Accepted Answer

The given deceleration rate is dv/dt = -2.5√v.
We need to find the time (t) taken for the object to come to rest, meaning its final speed (v_f) is 0 m/s, starting from an initial speed (v_i) of 6.25 m/s.

This is a differential equation that can be solved by separating variables and integrating.

**1. Separate variables:**
dv / √v = -2.5 dt

**2. Integrate both sides:**
∫ (1/√v) dv = ∫ -2.5 dt

**3. Set the limits of integration:**
For speed (v): from initial speed v_i = 6.25 m/s to final speed v_f = 0 m/s.
For time (t): from initial time t_i = 0 s to final time t_f = t.

∫ from 6.25 to 0 (v⁻¹/²) dv = ∫ from 0 to t (-2.5) dt

**4. Perform the integration:**
[ (v¹/²) / (1/2) ] from 6.25 to 0 = -2.5 [ t ] from 0 to t

[ 2√v ] from 6.25 to 0 = -2.5 (t - 0)

**5. Substitute the limits:**
(2√0) - (2√6.25) = -2.5t

0 - (2 × 2.5) = -2.5t

-5 = -2.5t

**6. Solve for t:**
t = -5 / -2.5
t = 2 s

Therefore, the time taken by the object to come to rest is 2 seconds.

The correct option is (a).

Question 17

Q17. A car moving along a straight highway with speed of 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long d

Accepted Answer

We are given the initial speed, final speed, and the stopping distance. We need to find the uniform retardation (negative acceleration) and the time taken to stop.

**Given:**
*   Initial speed (u) = 126 km h⁻¹
*   Final speed (v) = 0 m s⁻¹ (since the car is brought to a stop)
*   Distance (s) = 200 m

**1. Convert initial speed to m/s:**
u = 126 km/h = 126 × (1000 m / 3600 s) = 126 × (5/18) m/s = 7 × 5 m/s = 35 m/s

**2. Calculate retardation (a) using the third equation of motion:**
v² = u² + 2as
0² = (35 m/s)² + 2 × a × (200 m)
0 = 1225 m²/s² + 400a m
-400a = 1225
a = -1225 / 400 m/s²
a = -3.0625 m/s²

The retardation of the car is 3.0625 m/s² (the negative sign indicates deceleration).

**3. Calculate the time taken (t) using the first equation of motion:**
v = u + at
0 = 35 m/s + (-3.0625 m/s²) × t
-35 = -3.0625t
t = -35 / -3.0625 s
t = 11.42857 s
t ≈ 11.43 s

Alternatively, using the second equation of motion (if 'a' was not calculated first):
s = (u+v)/2 * t
200 = (35+0)/2 * t
200 = 17.5 * t
t = 200 / 17.5 = 11.42857 s ≈ 11.43 s

**Final Answer:**
The retardation of the car is **3.0625 m/s²**.
The time taken for the car to stop is approximately **11.43 s**.

Question 18

Q18. An object moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{dv}{dt} = -2.5\sqrt{v}$ where $v$ is the instantaneous speed of the object. The time taken by th

Accepted Answer

The time taken by the object to come to rest is 2.0 seconds.

Question 19

Q19. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 k

Accepted Answer

Let's first calculate the time taken for each part of the journey.
Distance to market, d = 2.5 km
Speed to market, v1 = 5 km/h
Time taken to reach market, t1 = d / v1 = 2.5 km / 5 km/h = 0.5 h = 30 min

Speed back home, v2 = 7.5 km/h
Time taken to return home, t2 = d / v2 = 2.5 km / 7.5 km/h = 1/3 h = 20 min

Total time for the round trip = t1 + t2 = 30 min + 20 min = 50 min

(a) Magnitude of average velocity = |Total Displacement| / Total Time
(b) Average speed = Total Distance / Total Time

**Case (i): Time interval 0 to 30 min**
During this interval, the man walks from home to the market.
Total time = 30 min = 0.5 h
Total distance = 2.5 km
Total displacement = 2.5 km (from home to market)

(a) Average velocity = Total Displacement / Total Time = 2.5 km / 0.5 h = 5 km/h
(b) Average speed = Total Distance / Total Time = 2.5 km / 0.5 h = 5 km/h

**Case (ii): Time interval 0 to 50 min**
During this interval, the man walks from home to market and then back home.
Total time = 50 min = 50/60 h = 5/6 h
Total distance = Distance to market + Distance back home = 2.5 km + 2.5 km = 5 km
Total displacement = 0 (since he starts and ends at home)

(a) Average velocity = Total Displacement / Total Time = 0 km / (5/6 h) = 0 km/h
(b) Average speed = Total Distance / Total Time = 5 km / (5/6 h) = 6 km/h

**Case (iii): Time interval 0 to 40 min**
This interval includes the time to reach the market (30 min) and an additional 10 min (40 min - 30 min = 10 min) of walking back home.

Time spent walking back home = 10 min = 10/60 h = 1/6 h
Distance covered while walking back home = v2 × (1/6 h) = 7.5 km/h × (1/6 h) = 1.25 km

Total time = 40 min = 40/60 h = 2/3 h
Total distance = Distance to market + Distance walked back = 2.5 km + 1.25 km = 3.75 km
Total displacement = Distance to market - Distance walked back = 2.5 km - 1.25 km = 1.25 km (from home, towards market)

(a) Average velocity = Total Displacement / Total Time = 1.25 km / (2/3 h) = 1.25 × (3/2) km/h = 1.875 km/h
(b) Average speed = Total Distance / Total Time = 3.75 km / (2/3 h) = 3.75 × (3/2) km/h = 5.625 km/h

Question 20

Q20. A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long d

Accepted Answer

Given:
Initial speed of the car, u = 126 km/h
Final speed of the car, v = 0 km/h (since it comes to a stop)
Stopping distance, s = 200 m

First, convert the initial speed from km/h to m/s:
u = 126 km/h = 126 × (1000 m / 3600 s) = 126 × (5/18) m/s = 7 × 5 m/s = 35 m/s

**Part 1: Calculate the retardation (acceleration 'a')**
We can use the third equation of motion:
v² = u² + 2as
0² = (35 m/s)² + 2 × a × 200 m
0 = 1225 m²/s² + 400a m
-400a = 1225
a = -1225 / 400 m/s²
a = -3.0625 m/s²

The negative sign indicates retardation (deceleration). Therefore, the retardation of the car is 3.0625 m/s².

**Part 2: Calculate the time taken for the car to stop (time 't')**
We can use the first equation of motion:
v = u + at
0 = 35 m/s + (-3.0625 m/s²) × t
3.0625t = 35
t = 35 / 3.0625 s
t = 11.42857 s

Rounding to a reasonable number of significant figures, t ≈ 11.43 s.

Thus, the retardation of the car is 3.06 m/s² and it takes approximately 11.43 seconds for the car to stop.

Question 21

Q21. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overt

Accepted Answer

Let's denote the initial distance between the driver of A and the guard of B as 'x'.

Given:
Length of train A, L_A = 400 m
Length of train B, L_B = 400 m
Initial speed of train A, v_A = 72 km/h
Initial speed of train B, v_B = 72 km/h
Acceleration of train B, a_B = 1 m/s²
Time taken to overtake, t = 50 s

First, convert speeds from km/h to m/s:
v_A = 72 km/h = 72 × (5/18) m/s = 4 × 5 m/s = 20 m/s
v_B = 72 km/h = 20 m/s

Consider the motion of train B relative to train A.
Initial relative velocity, u_rel = v_B - v_A = 20 m/s - 20 m/s = 0 m/s
Relative acceleration, a_rel = a_B - a_A. Since train A moves with uniform speed, a_A = 0 m/s².
So, a_rel = 1 m/s² - 0 m/s² = 1 m/s²

For train B to just pass train A, the guard of B must pass the driver of A. This means the total relative displacement covered by train B with respect to train A must be equal to the initial separation between the driver of A and the guard of B, plus the length of train A, plus the length of train B.

Let the original distance between the driver of A and the guard of B be 'd'.
When the guard of B just passes the driver of A, the total relative distance covered by B with respect to A is:
S_rel = (Initial distance between driver of A and guard of B) + (Length of A) + (Length of B)
S_rel = d + L_A + L_B
S_rel = d + 400 m + 400 m = d + 800 m

Using the equation of motion for relative displacement:
S_rel = u_rel × t + (1/2) × a_rel × t²

Substitute the values:
d + 800 = (0 m/s) × 50 s + (1/2) × (1 m/s²) × (50 s)²
d + 800 = 0 + (1/2) × 1 × 2500
d + 800 = 1250
d = 1250 - 800
d = 450 m

Therefore, the original distance between the driver of A and the guard of B was 450 m.

Question 22

Q22. On a two-lane road, car A is moving with a speed of 36 km/h. Two cars B and C approach car A from opposite directions with a speed of 54 km/h each. At a certain instant, when the d

Accepted Answer

Let's denote the speeds and distances:
Speed of car A, v_A = 36 km/h
Speed of car B, v_B = 54 km/h
Speed of car C, v_C = 54 km/h
Initial distance AB = 1 km = 1000 m
Initial distance AC = 1 km = 1000 m

First, convert all speeds to m/s:
v_A = 36 km/h = 36 × (5/18) m/s = 10 m/s
v_B = 54 km/h = 54 × (5/18) m/s = 15 m/s
v_C = 54 km/h = 15 m/s

**Step 1: Determine the time available for B to overtake A**
Car C is approaching car A from the opposite direction. An accident will occur if C meets A before B overtakes A.
Relative speed of C with respect to A (v_CA) = v_C + v_A (since they are moving in opposite directions)
v_CA = 15 m/s + 10 m/s = 25 m/s

Distance C needs to cover to meet A = AC = 1000 m
Time taken for C to meet A (t_C) = Distance / Relative Speed = 1000 m / 25 m/s = 40 s

This means car B must overtake car A within 40 seconds to avoid an accident.
So, the maximum time available for B to overtake A is t_max = 40 s.

**Step 2: Calculate the minimum acceleration required for car B**
Car B needs to overtake car A. This means car B must cover the initial distance AB (1000 m) relative to A.
Consider the motion of car B relative to car A.
Initial relative velocity of B with respect to A (u_BA) = v_B - v_A (since they are moving in the same direction)
u_BA = 15 m/s - 10 m/s = 5 m/s

Relative displacement required for B to overtake A (S_BA) = 1000 m
Time available (t) = 40 s
Let the minimum acceleration of car B be 'a_B'. Car A has no acceleration (a_A = 0).
Relative acceleration of B with respect to A (a_BA) = a_B - a_A = a_B - 0 = a_B

Using the equation of motion for relative displacement:
S_BA = u_BA × t + (1/2) × a_BA × t²

Substitute the values:
1000 = (5 m/s) × 40 s + (1/2) × a_B × (40 s)²
1000 = 200 + (1/2) × a_B × 1600
1000 = 200 + 800 a_B
800 a_B = 1000 - 200
800 a_B = 800
a_B = 800 / 800 m/s²
a_B = 1 m/s²

Therefore, the minimum acceleration of car B required to avoid an accident is 1 m/s².

Question 23

Q23. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h⁻¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5

Accepted Answer

To solve this problem, we first calculate the time taken for each part of the journey.

**Given Data:**
*   Distance from home to market (d) = 2.5 km
*   Speed from home to market (v₁) = 5 km/h
*   Speed from market to home (v₂) = 7.5 km/h

**Calculations for individual journey segments:**
1.  **Time taken to reach the market (t₁):**
    t₁ = Distance / Speed = 2.5 km / 5 km/h = 0.5 h = 30 minutes.
2.  **Time taken to return home from the market (t₂):**
    t₂ = Distance / Speed = 2.5 km / 7.5 km/h = 1/3 h = 20 minutes.

Now, we calculate average velocity and average speed for the specified time intervals.

**(i) Time interval: 0 to 30 min**
*   At t = 30 min, the man has just reached the market.
*   **Total distance covered:** 2.5 km
*   **Total displacement:** 2.5 km (assuming home is the origin and market is in the positive direction)
*   **Total time taken:** 30 min = 0.5 h

**(a) Magnitude of average velocity:**
    Average velocity = Total Displacement / Total Time
    Average velocity = 2.5 km / 0.5 h = 5 km/h

**(b) Average speed:**
    Average speed = Total Distance / Total Time
    Average speed = 2.5 km / 0.5 h = 5 km/h

**(ii) Time interval: 0 to 50 min**
*   At t = 30 min, the man reaches the market.
*   From t = 30 min to t = 50 min (i.e., for 20 min), he walks back towards home.
*   Distance covered while returning in 20 min = Speed × Time = 7.5 km/h × (20/60 h) = 7.5 km/h × (1/3 h) = 2.5 km.

*   **Total distance covered:** 2.5 km (to market) + 2.5 km (back home) = 5 km
*   **Total displacement:** 2.5 km (positive) - 2.5 km (negative) = 0 km
*   **Total time taken:** 50 min = 50/60 h = 5/6 h

**(a) Magnitude of average velocity:**
    Average velocity = Total Displacement / Total Time
    Average velocity = 0 km / (5/6 h) = 0 km/h

**(b) Average speed:**
    Average speed = Total Distance / Total Time
    Average speed = 5 km / (5/6 h) = 5 × 6 / 5 km/h = 6 km/h

**(iii) Time interval: 0 to 40 min**
*   At t = 30 min, the man reaches the market.
*   From t = 30 min to t = 40 min (i.e., for 10 min), he walks back towards home.
*   Distance covered while returning in 10 min = Speed × Time = 7.5 km/h × (10/60 h) = 7.5 km/h × (1/6 h) = 1.25 km.

*   **Total distance covered:** 2.5 km (to market) + 1.25 km (back home) = 3.75 km
*   **Total displacement:** 2.5 km (positive) - 1.25 km (negative) = 1.25 km
*   **Total time taken:** 40 min = 40/60 h = 2/3 h

**(a) Magnitude of average velocity:**
    Average velocity = Total Displacement / Total Time
    Average velocity = 1.25 km / (2/3 h) = 1.25 × 3 / 2 km/h = 3.75 / 2 km/h = 1.875 km/h

**(b) Average speed:**
    Average speed = Total Distance / Total Time
    Average speed = 3.75 km / (2/3 h) = 3.75 × 3 / 2 km/h = 11.25 / 2 km/h = 5.625 km/h

Question 24

Q24. A body travelling in a straight line with a uniform acceleration 'a'. At some instant its velocity is 10 m/s. After 5 s its velocity is 20 m/s. Find the velocity at 3 s before that

Accepted Answer

Let's denote the instant when the velocity is 10 m/s as t = 0.

**Given Data:**
*   Velocity at t = 0 (u) = 10 m/s
*   Velocity at t = 5 s (v) = 20 m/s
*   Acceleration (a) is uniform.

**Step 1: Calculate the uniform acceleration (a).**
We use the first equation of motion: v = u + at
Substituting the given values:
20 m/s = 10 m/s + a × 5 s
20 - 10 = 5a
10 = 5a
a = 10 / 5 m/s²
a = 2 m/s²

**Step 2: Calculate the velocity at 3 s before that instant.**
'That instant' refers to t = 0. So, 3 s before that instant means at t = -3 s.
Let the velocity at t = -3 s be v'.
Using the first equation of motion again: v' = u + at
Here, u is the velocity at t=0 (10 m/s), and t is -3 s.
v' = 10 m/s + (2 m/s²) × (-3 s)
v' = 10 m/s - 6 m/s
v' = 4 m/s

Thus, the velocity of the body at 3 s before the instant when its velocity was 10 m/s is 4 m/s.

Chapter 3: Motion in a Straight Line