RBSE Solutions Class 9 Science Chapter 16 Road Safety Education

 

Question 1. Why there is more chance of an accident during over speed compare to slow speed?
Answer: When a vehicle moves at a very high speed, the driver finds it difficult to maintain full control. If something unexpected happens, like needing to stop quickly or swerve, the high speed makes it much harder to react and steer safely. This loss of control or confusion due to sudden events often leads to accidents. Driving at a reasonable speed gives the driver more time to respond to dangers.
In simple words: Faster driving makes it harder to control the car and react to problems, which causes more accidents.

🎯 Exam Tip: Always relate high speed to reduced control and increased reaction time, which are key factors in accident causation.

 

Question 2. A heavy truck and car, both are moving with same speed and after a mutual head-on collision, they come to rest. If the rate of change of momentum is the same for both of them, then why there is more damage to the car in comparison to the truck?
Answer: Momentum (\(p\)) is calculated as mass times velocity (\(p = mv\)). Kinetic energy (\(K\)) is given by \(K = \frac{1}{2}mv^2\). We can also write kinetic energy in terms of momentum as \(K = \frac{p^2}{2m}\). If both the truck and the car have the same rate of change of momentum (meaning the impulse they experience is the same), then the car, being much lighter (having a smaller mass \(m\)), will experience a greater change in kinetic energy compared to the truck. This larger change in kinetic energy for the car translates to more work done on it during the collision, causing more damage. The energy transferred during the collision is absorbed by the deformation of the vehicles.
In simple words: Even if the push (change in momentum) is the same for both, the lighter car gets more damage because it absorbs more energy during the crash compared to the heavier truck.

🎯 Exam Tip: Remember that kinetic energy is related to momentum by \( K = \frac{p^2}{2m} \). For a given momentum change, a smaller mass means a larger kinetic energy change and thus more damage.

 

Question 3. Why F - I car race tracks are specially designed?
Answer: F-1 (Formula 1) car race tracks are specially designed to help drivers manage speed, braking, and steering safely. These tracks have specific features like banking in turns, wide run-off areas, and engineered surfaces that allow drivers to maintain control at extremely high speeds, reduce stopping distances, and optimize their reaction times. This design enhances both performance and safety, as the high speeds involved in F-1 racing demand precise conditions to prevent accidents.
In simple words: F-1 tracks are built in a special way to help racing drivers control their fast cars better, stop safely, and react quickly to avoid accidents.

🎯 Exam Tip: When discussing track design, focus on how specific features (banking, run-off areas) directly impact vehicle performance and safety at high speeds.

 

Question 4. Observe the vehicle moving ahead of you. When it crosses the stationary object like a tree, pole etc., then count the time (in seconds taken by you to cross the objects. If you reach before, then decrease the speed and if you reach after, then increase the speed. By which instrument will you count the seconds?
Answer: To accurately count the seconds when performing this observation and adjusting speed, one would use a stopwatch. A stopwatch helps measure short periods of time with good precision, which is essential for this exercise in judging distance and speed relative to another vehicle and stationary objects.
In simple words: You use a stopwatch to count the seconds for this driving exercise.

🎯 Exam Tip: For questions asking about measuring time, the most common and direct answer is usually a stopwatch.

 

Question 5. Complete the following table:
How these distances vary on wet road?
Answer:

In the second situation, the total stopping distance (54m) is derived from the reaction distance and following distance time:
Total stopping distance = Reaction distance \( \times \) following distance time
\( 54 = x \times 3 \)
\( \implies x = \frac{54}{3} \)
\( \implies x = 18m \)

In the third situation, the total stopping distance (108m) is calculated as:
Total stopping distance = 27 \( \times \) 4 = 108 m

On a wet road, the reaction distance will increase because it takes longer for the tires to grip, and thus the total stopping distance will also increase significantly. Wet roads reduce friction, making it harder to slow down.
In simple words: The table shows how stopping distance increases with speed. On wet roads, it takes longer to react and stop because there is less grip, so all these distances would become much bigger.

🎯 Exam Tip: When completing tables, look for patterns or formulas provided. For wet roads, remember that reduced friction always leads to longer stopping and reaction distances.

 

Road Safety Education Additional Question Solved

 

Question 1. On which factors, stopping distance depends?
Answer: Stopping distance, which is the total distance a vehicle travels from the moment the driver notices a hazard until the vehicle comes to a complete stop, depends on several important factors:
1. Condition of roads (e.g., wet, dry, icy, uneven).
2. Weather conditions (e.g., rain, snow, fog, clear).
3. The condition of the vehicle's tyres and brakes.
All these factors influence the friction between the tires and the road, as well as the vehicle's ability to decelerate.
In simple words: How far a car goes before it stops depends on if the road is good or bad, the weather, and how good the car's tires and brakes are.

🎯 Exam Tip: When listing factors for stopping distance, always consider both environmental conditions (road, weather) and vehicle mechanics (tires, brakes).

 

Road Safety Education Textbook Questions Solved

 

Question 1. Why threads are made on Tyres?
Answer: Threads, also known as treads or grooves, are made on the surface of tires to increase the friction between the tire and the road. This improved friction helps the vehicle grip the road better, especially in wet conditions by channeling water away. Without these treads, tires would easily slip, leading to a loss of control and increasing the risk of accidents.
In simple words: Tire treads help tires grip the road and stop slipping, especially when it's wet, by increasing friction.

🎯 Exam Tip: The main purpose of tire treads is to enhance friction and prevent slipping, particularly by dispersing water from under the tire.

 

Question 2. Why there are different threads on a different type of tyres. How they are helpful, in case of accidents.
Answer: Different types of tires have different tread patterns because they are designed for specific driving conditions and purposes. For example, summer tires have different treads than winter tires, which are designed to bite into snow and ice. These varying tread patterns create different amounts and types of friction, which helps prevent cars from slipping in various situations, thus reducing the chance of accidents. For instance, deeper, wider grooves are effective in wet conditions, while more sipes (small cuts) help grip icy roads.
In simple words: Different tires have different patterns to work best in certain weather or road conditions, helping to stop the car from slipping and causing crashes.

🎯 Exam Tip: Emphasize that tread design is specialized for specific conditions (wet, snow, dry) to optimize friction and prevent slipping.

 

Question 3. Why air in tyres should neither be excessive nor be less than fixed range?
Answer: The air pressure in tires must be kept within a specific range for safety and efficiency. If the air pressure is too high (excessive), the tire can become overinflated, making it more rigid and increasing the risk of a blowout due to high internal pressure. On the other hand, if the air pressure is too low, the tire becomes underinflated, which increases its rolling resistance and causes more fuel consumption. Low pressure also makes the tire sidewalls flex too much, building up heat and potentially leading to a tire failure or a puncture. Maintaining the correct pressure helps the tire contact the road properly, ensuring even wear and good handling.
In simple words: Too much air in tires can make them burst, while too little air can use more fuel and cause punctures. So, the right amount is needed.

🎯 Exam Tip: Remember the two main dangers: overinflation leads to blowouts, and underinflation leads to increased wear, poor fuel economy, and potential punctures.

 

Question 4. Why we use fuel oil in vehicles?
Answer: Fuel oil, such as petrol or diesel, is used in vehicles because it powers the engine. The engine operates by converting the chemical energy stored in the fuel into mechanical energy. This mechanical energy then drives the wheels, allowing the vehicle to move. This conversion happens through a controlled combustion process within the engine cylinders.
In simple words: We use fuel oil to make the engine run. The fuel gives the engine energy to move the vehicle.

🎯 Exam Tip: Focus on the energy conversion: fuel contains chemical energy, which the engine converts into mechanical energy to move the vehicle.

 

Question 6. During the collision of a truck and a car, on which body more force works? Explain your answer.
Answer: According to Newton's Third Law of Motion, during a collision between a truck and a car, both bodies experience an equal and opposite force. This means that the force exerted by the car on the truck is exactly the same magnitude as the force exerted by the truck on the car. However, because the car has a much smaller mass than the truck, this equal force causes a greater acceleration (and thus greater impact/damage) to the car. In terms of momentum, if \( M_1 \) is the mass of the truck and \( m_2 \) is the mass of the car, and their initial velocities are \( u_1 \) and \( u_2 \) respectively, and final velocities are \( v_1 \) and \( v_2 \), then according to the law of conservation of momentum: \( M_1u_1 + m_2u_2 = M_1v_1 + m_2v_2 \). Although the change in momentum is equal and opposite for both, the car experiences a larger acceleration and typically suffers more damage because its mass is much smaller.
In simple words: Both the car and the truck feel the same amount of force during a crash. But because the car is much lighter, that same force causes more damage to the car.

🎯 Exam Tip: Always remember Newton's Third Law: forces in a collision are always equal and opposite. The *effect* of that force (damage, acceleration) depends on the mass.

 

Question 7. When you are moving on a circular path, then what type of force work on your body.
Answer: When you are moving on a circular path, a centripetal force acts on your body. This force is directed towards the center of the circular path and is necessary to keep you moving in a curve rather than in a straight line. Without it, you would continue in a tangent to the circle.
In simple words: A centripetal force pulls you towards the middle of the circle when you move in a curve.

🎯 Exam Tip: Centripetal force is crucial for circular motion and is always directed towards the center of the circle.

 

Question 8. There is a restriction of overspend on the circular or spiral path? Why.
Answer: There is a restriction on overspending (or exceeding speed limits) on circular or spiral paths because going too fast can cause the vehicle to lose traction and possibly topple over. For a circular path with radius \(r\), the maximum safe speed (\(v\)) depends on the coefficient of static friction (\(\mu_s\)) between the tires and the road, and gravity (\(g\)). The formula for maximum safe speed on an unbanked curve is approximately \( v = \sqrt{r g \mu_s} \). If the vehicle's speed is greater than this maximum value, the necessary centripetal force cannot be provided by friction alone, leading to skidding or, in extreme cases, the vehicle overturning.
In simple words: You cannot drive too fast on curved roads because the car might slide off or tip over. The friction from the road can only hold the car up to a certain speed.

🎯 Exam Tip: Link high speed on curves to insufficient centripetal force from friction, leading to skidding or toppling. Mentioning the role of friction is key.

 

Question 9. For safe driving, brakes are very important. Why any vehicle stop after applying brakes? Describe the working methods of brakes?
Answer: Brakes are very important for safe driving because they allow a vehicle to slow down and stop reliably. A vehicle stops after applying brakes because the braking system generates a strong frictional force that acts in the opposite direction of the vehicle's motion. This opposing force converts the vehicle's kinetic energy into heat, causing it to decelerate until it comes to rest.

Most vehicles use hydraulic brakes. Here's how they work:
1. **Master Cylinder:** When the driver presses the brake pedal, a piston (P) in the master cylinder (M) moves. This cylinder is filled with brake fluid (hydraulic oil).
2. **Pressure Distribution:** According to Pascal's Principle, the pressure created in the brake fluid is distributed equally through a tube (T) to all the wheel cylinders (C).
3. **Wheel Cylinders:** Each wheel cylinder contains pistons (P1 and P2) that are connected to brake shoes (S1 and S2).
4. **Friction Application:** The pressure from the fluid pushes these pistons outwards, forcing the brake shoes against the rotating wheel drums or discs. This action creates a strong frictional force that slows down and stops the wheels.
This entire system ensures that a small force on the brake pedal can generate a large braking force at the wheels, making stopping efficient and safe.
In simple words: Brakes make a car stop by creating friction that works against the car's movement. When you press the pedal, fluid pushes pads against the wheels, slowing them down.

🎯 Exam Tip: When describing brake function, highlight the role of friction, the conversion of kinetic energy to heat, and Pascal's principle in hydraulic systems.

 

Question 10. What is Airbag? During accident of the vehicle, how they provide safety. Explain.
Answer: An airbag is an important safety device in a vehicle. It is designed to inflate extremely quickly during a collision, creating a soft cushion between the occupant and the hard surfaces of the vehicle, like the steering wheel or dashboard. Airbags work by rapidly releasing gas upon impact, which slows down the occupant's movement in a crash. This helps to spread the force of the impact over a larger area of the body and reduces the risk of severe injuries, particularly to the head and chest. However, the effectiveness of airbags is significantly reduced if the seat belt is not worn, as the occupant might move out of position before the airbag fully inflates.
In simple words: Airbags are cushions that pop out in a crash to stop you from hitting hard parts of the car. They protect you by spreading out the impact force, but they work best when you also wear your seat belt.

🎯 Exam Tip: Explain that airbags work by creating a cushion and increasing the impact time, thus reducing the force on the occupant, and always mention the importance of seat belts.

 

Road Safety Education Textbook Questions Solved

 

Question 1. A loaded truck of 2500 kg is moving at a speed of 30 km/h. On the way he stops at a factory and 500 kg of extra weight is placed on it and again he moves with the speed of 30 km/h. Calculate the change in kinetic energy of the truck and also work done by it. Also find out the inference about the change in kinetic energy, due to the increase or decrease of mass. What do you think, is there any change in momentum also. If yes, then how much?
Answer:First, convert the speed from km/h to m/s:
Speed \( = 30 \, \text{km/h} = 30 \times \frac{5}{18} \, \text{m/s} = \frac{25}{3} \, \text{m/s} \)

**Case I: Initial state**
Mass of truck (\( m_1 \)) = 2500 kg
Speed of truck (\( v \)) = \( \frac{25}{3} \, \text{m/s} \)
Initial Kinetic energy (\( K_i \)) \( = \frac{1}{2} m_1 v^2 \)
\( = \frac{1}{2} \times 2500 \times \left( \frac{25}{3} \right)^2 \)
\( = \frac{1}{2} \times 2500 \times \frac{625}{9} \)
\( = \frac{1250 \times 625}{9} = \frac{781250}{9} \, \text{Joules} \)

**Case II: Final state (after adding weight)**
Mass of truck (\( m_2 \)) = 2500 kg + 500 kg = 3000 kg
Speed of truck (\( v \)) = \( \frac{25}{3} \, \text{m/s} \)
Final Kinetic energy (\( K_f \)) \( = \frac{1}{2} m_2 v^2 \)
\( = \frac{1}{2} \times 3000 \times \left( \frac{25}{3} \right)^2 \)
\( = \frac{1}{2} \times 3000 \times \frac{625}{9} \)
\( = \frac{1500 \times 625}{9} = \frac{937500}{9} \, \text{Joules} \)
\( = \frac{312500}{3} \, \text{Joules} \)

**Change in kinetic energy (\( \Delta K \))**
\( \Delta K = K_f - K_i \)
\( = \frac{937500}{9} - \frac{781250}{9} \)
\( = \frac{937500 - 781250}{9} \)
\( = \frac{156250}{9} \, \text{Joules} \)
\( \approx 17361 \, \text{Joules} \)

**Work done**
The work done by the truck is equal to the change in its kinetic energy.
Work done \( = \Delta K = \frac{156250}{9} \, \text{Joules} \approx 17361 \, \text{Joules} \)

**Inference about change in kinetic energy due to increase or decrease of mass:**
When the mass of the truck increased while its speed remained constant, its kinetic energy also increased. This shows that kinetic energy is directly proportional to mass. If the mass were to decrease, the kinetic energy would also decrease. Therefore, changes in mass directly affect kinetic energy when speed is constant.

**Change in momentum:**
Yes, there is a change in momentum. Momentum (\(p\)) is given by \(p = mv\). Since the mass of the truck increased from 2500 kg to 3000 kg while its velocity remained constant, its momentum also increased.
Initial momentum (\( p_i \)) \( = m_1 v = 2500 \times \frac{25}{3} = \frac{62500}{3} \, \text{kg m/s} \)
Final momentum (\( p_f \)) \( = m_2 v = 3000 \times \frac{25}{3} = 25000 \, \text{kg m/s} \)
Change in momentum (\( \Delta p \)) \( = p_f - p_i = 25000 - \frac{62500}{3} \)
\( = \frac{75000 - 62500}{3} = \frac{12500}{3} \, \text{kg m/s} \)
\( \approx 4166.67 \, \text{kg m/s} \)
The momentum increased by approximately \( 4166.67 \, \text{kg m/s} \).
In simple words: When the truck carried more weight, its kinetic energy and momentum both increased, even though its speed stayed the same. The extra weight meant more energy was needed to move it, and it also had a bigger push if it hit something.

🎯 Exam Tip: Remember to convert units (km/h to m/s) at the start of physics problems. Clearly state the formulas for kinetic energy and momentum and how they depend on mass and velocity.

 

Question 2. When there is a dynamic collision between two vehicles, how kinetic energy change into heat energy?
Answer: During a dynamic collision between two vehicles, especially if they stick together (an inelastic collision), a significant portion of the kinetic energy they possessed before the crash is transformed into other forms of energy. This conversion primarily occurs as heat energy, sound energy, and the energy used to deform the colliding bodies (causing damage). According to the work-energy principle, the work done in deforming the vehicles and generating heat during the impact accounts for the loss of mechanical kinetic energy in the system.
In simple words: In a crash where cars stick together, the energy of their movement turns into heat, sound, and bending the metal of the cars.

🎯 Exam Tip: For inelastic collisions, remember that kinetic energy is not conserved; it's converted into heat, sound, and deformation energy due to the work done during impact.

 

Question 3. Differentiate between Elastic and Non-elastic collision?
Answer:

The key difference lies in the conservation of kinetic energy: it is conserved in elastic collisions but not in non-elastic ones.
In simple words: In elastic crashes, the energy of movement stays the same, but in non-elastic crashes, some movement energy changes into heat or damages the objects. In both, the total push (momentum) always stays the same.

🎯 Exam Tip: Clearly state that momentum is always conserved in both types of collisions, but kinetic energy is only conserved in elastic collisions. Energy conversion into heat/deformation is characteristic of inelastic collisions.

 

Road Safety Education Textbook Questions Solved

 

Question 1. Draw the symbol of NO HORN ZONE.
(a) In road symbols, what does circle signify?
(b) What does red circle signify?
Answer:
(a) In road symbols, a circular shape generally signifies a "regulatory" or "mandatory" instruction. These symbols tell drivers what they *must* or *must not* do.
(b) A red circle with a diagonal line (like in the "No Horn Zone" symbol) indicates a "prohibition" or "restriction." It tells drivers that a particular action is forbidden or not allowed.
In simple words: A circle on a road sign usually means it's a rule you must follow. If it's a red circle with a line through it, it means "don't do this."

🎯 Exam Tip: For road signs, remember that circles are regulatory (rules), and red circles with a diagonal line specifically indicate prohibitions (forbidden actions).

 

Question 2. What is the audible range for the human being?
Answer: The audible range for human beings, which is the range of sound frequencies they can typically hear, extends from approximately 20 Hertz (Hz) to 20,000 Hertz (Hz). Sounds below 20 Hz are infrasound, and those above 20,000 Hz are ultrasound, neither of which is audible to humans. This range can vary slightly between individuals and tends to decrease with age.
In simple words: Humans can hear sounds that are between 20 Hertz and 20,000 Hertz.

🎯 Exam Tip: Remember the specific range for human hearing (20 Hz to 20,000 Hz) and note that it can change with age.

 

Question 3. How is noise measured?
Answer: Noise is measured in units called decibels (dB). The decibel scale is a logarithmic scale, which means that a small increase in decibel value represents a large increase in sound intensity. A higher decibel reading indicates a louder sound, and prolonged exposure to high decibel levels can be harmful to human hearing.
In simple words: Noise is measured in decibels (dB), which tells us how loud a sound is.

🎯 Exam Tip: The key unit for measuring noise is decibels (dB); remember that it indicates sound intensity or loudness.