set 13k

Q1 [3 Marks]: Define 'Reaction Time'. On what human and environmental factors does the reaction time of a driver depend?

Part A: Definition
Reaction Time is the time interval between a person observing a situation (like an obstacle on the road) and taking the appropriate physical action (like pressing the brake pedal). During this time, the vehicle continues to move at its initial uniform velocity.

Part B: Factors Influencing Reaction Time
Reaction time is not constant and varies based on:

• Human Factors: The driver's age, level of fatigue, state of intoxication (alcohol/drugs), visual acuity, and level of distraction (e.g., using a phone).
• Environmental Factors: Visibility conditions (fog, night time, heavy rain) and the complexity of the traffic situation (which increases cognitive processing time).

Q2 [3 Marks]: A car moving at $60\text{ km/h}$ is brought to a halt in $20\text{ m}$. If the same car is moving at $120\text{ km/h}$, calculate the minimum stopping distance assuming the same braking force.

Step 1: Establish the mathematical relationship.
From the third equation of motion ($v^2 = u^2 - 2as$), when a car comes to a halt, final velocity $v = 0$.

$$0 = u^2 - 2as \Rightarrow s = \frac{u^2}{2a}$$

Since the braking force is the same, the deceleration $a$ is constant. Therefore, stopping distance $s$ is directly proportional to the square of the initial velocity $u^2$ ($s \propto u^2$).

Step 2: Use proportionality to find the new distance.
Let initial state be $u_1 = 60\text{ km/h}$ and $s_1 = 20\text{ m}$.
Let final state be $u_2 = 120\text{ km/h}$ and $s_2$ be unknown.

$$\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2$$

$$\frac{s_2}{20} = \left(\frac{120}{60}\right)^2 = (2)^2 = 4$$

Step 3: Calculate $s_2$.

$$s_2 = 20 \times 4$$

$$\text{Minimum Stopping Distance } = 80\text{ m}$$

Pro-Tip: When speed doubles, stopping distance quadruples! This is why speeding is so dangerous.

Q3 [4 Marks]: A driver takes $0.20\text{ s}$ to apply the brakes after seeing an obstacle. If he is driving at $54\text{ km/h}$ and the brakes produce a retardation of $6.0\text{ m/s}^2$, calculate the total distance traveled before stopping.

Step 1: Convert velocity to SI units.

$$u = 54\text{ km/h} = 54 \times \frac{5}{18} = 15\text{ m/s}$$

Step 2: Calculate the Reaction Distance ($d_r$).
During the reaction time ($t_r = 0.20\text{ s}$), the car continues to move at a constant speed of $15\text{ m/s}$ before the brakes are engaged.

$$d_r = u \times t_r = 15 \times 0.20$$

$$d_r = 3.0\text{ m}$$

Step 3: Calculate the Braking Distance ($d_b$).
Now the brakes are applied. $u = 15\text{ m/s}$, final velocity $v = 0$, and retardation $a = 6.0\text{ m/s}^2$.

$$v^2 = u^2 - 2a(d_b) \Rightarrow 0 = 15^2 - 2(6.0)d_b$$

$$12d_b = 225 \Rightarrow d_b = \frac{225}{12}$$

$$d_b = 18.75\text{ m}$$

Step 4: Calculate Total Stopping Distance.

$$\text{Total Distance} = \text{Reaction Distance} + \text{Braking Distance}$$

$$\text{Total Distance} = 3.0 + 18.75$$

$$\text{Total Distance Traveled } = 21.75\text{ m}$$

Q4 [5 Marks]: Two trains approach each other on the same track, both moving at $72\text{ km/h}$. When they are $2\text{ km}$ apart, both drivers simultaneously apply their brakes, which produce a retardation of $0.2\text{ m/s}^2$. Will there be a collision? Mathematically justify your answer.

Step 1: Convert values to SI units.
Initial velocity of both trains, $u = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$.
Initial separation distance $= 2\text{ km} = 2000\text{ m}$.
Retardation for both trains, $a = 0.2\text{ m/s}^2$.

Step 2: Calculate the stopping distance for ONE train.
We need to find how far one train travels before coming to a complete stop ($v = 0$).

$$v^2 = u^2 - 2as$$

$$0 = (20)^2 - 2(0.2)s$$

$$0.4s = 400$$

$$s = \frac{400}{0.4} = 1000\text{ m}$$

Each train will travel exactly $1000\text{ m}$ (or $1\text{ km}$) before stopping.

Train Stopping Distance Analysis

Step 3: Analyze the total distance required.
For the trains to avoid a collision, the sum of their stopping distances must be less than or equal to their initial separation.

$$\text{Total Distance Required} = s_1 + s_2 = 1000\text{ m} + 1000\text{ m} = 2000\text{ m}$$

Since their initial separation is exactly $2000\text{ m}$, the trains will come to rest at the exact same point at the exact same time without crashing into each other.

$$\text{NO Collision. They will just barely touch (or stop nose-to-nose).}$$