Set 9k

Q1 [3 Marks]: What does the slope of a relative position-time graph represent? What does it mean if the lines for two objects on this graph are parallel?

Part 1: Meaning of the Slope
The slope of a standard position-time ($x-t$) graph gives the velocity of the object. Therefore, the slope of a relative position-time graph ($x_{rel}$ vs $t$) represents the relative velocity of one object with respect to the other ($v_{rel} = v_A - v_B$).

Part 2: Meaning of Parallel Lines
If the $x-t$ lines for two objects are completely parallel, it means their slopes are identical. Since slope equals velocity, both objects are moving with the exact same velocity (same speed and same direction). Because their relative velocity is zero, the distance between them remains constant over time.

Q2 [3 Marks]: Two trains A and B of length $400\text{ m}$ each are moving on two parallel tracks with a uniform speed of $72\text{ km/h}$ in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by $1\text{ m/s}^2$. If after $50\text{ s}$, the guard of B just brushes past the driver of A, what was the original distance between them?

Step 1: Initial state & Conversions.
Initial velocity of A, $u_A = 72\text{ km/h} = 20\text{ m/s}$.
Initial velocity of B, $u_B = 72\text{ km/h} = 20\text{ m/s}$.
Since they are moving at the same speed in the same direction, their initial relative velocity is zero: $u_{rel} = u_B - u_A = 0$.

Step 2: Relative acceleration and time.
Train A has no acceleration ($a_A = 0$). Train B accelerates at $a_B = 1\text{ m/s}^2$.
Relative acceleration of B w.r.t A: $a_{rel} = a_B - a_A = 1\text{ m/s}^2$.
Time taken for the overtake event: $t = 50\text{ s}$.

Step 3: Calculate total relative distance covered by B.
Using the kinematic equation in the relative frame:

$$S_{rel} = u_{rel}t + \frac{1}{2}a_{rel}t^2$$

$$S_{rel} = (0)(50) + \frac{1}{2}(1)(50)^2 = \frac{1}{2}(2500) = 1250\text{ m}$$

This means Train B moved $1250\text{ m}$ forward relative to Train A.

Train Overtaking Setup (Relative Frame)

Step 4: Analyze the physical distances.
Look at the front of Train B (Driver B). For the back of Train B (Guard B) to align with the front of Train A (Driver A), Driver B must travel across:

• The initial gap between the trains ($D$)
• The entire length of Train A ($400\text{ m}$)
• Its own length, so its tail can pass the driver ($400\text{ m}$)

$$S_{rel} = D + L_A + L_B$$

$$1250 = D + 400 + 400$$

$$1250 = D + 800$$

$$\text{Original Distance } (D) = 450\text{ m}$$

Q3 [4 Marks]: A man running at $8\text{ km/h}$ on a straight road is ahead of a bus traveling at $40\text{ km/h}$. If the bus is $1\text{ km}$ behind the man, how long will it take for the bus to overtake him? Provide the answer in minutes.

Step 1: Calculate the relative velocity.
Both the man and the bus are moving in the same direction. The relative velocity of the bus with respect to the man ($v_{rel}$) is the difference in their speeds.

$$v_{rel} = v_{bus} - v_{man} = 40 - 8 = 32\text{ km/h}$$

This means the bus is closing the gap at a rate of $32\text{ km/h}$.

Step 2: Calculate time taken in hours.
The initial relative distance (gap) is $D = 1\text{ km}$.

$$t = \frac{\text{Distance}}{v_{rel}} = \frac{1\text{ km}}{32\text{ km/h}} = \frac{1}{32}\text{ hours}$$

Step 3: Convert time to minutes.
Since $1\text{ hour} = 60\text{ minutes}$:

$$t = \frac{1}{32} \times 60\text{ mins} = \frac{60}{32}\text{ mins}$$

$$t = \frac{15}{8} = 1.875\text{ mins}$$

$$\text{Time to overtake } = 1.875\text{ minutes}$$

Q4 [5 Marks]: 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 $20\text{ km/h}$ in the direction A to B notices that a bus goes past him every $18\text{ min}$ in the direction of his motion, and every $6\text{ min}$ in the opposite direction. What is the period $T$ and the speed of the buses?

Step 1: Establish the variables.
Let the uniform speed of the buses be $V\text{ km/h}$.
Speed of the cyclist, $v = 20\text{ km/h}$.
Since buses leave every $T$ minutes, the physical distance between two consecutive buses on the road is $VT$.

Step 2: Formulate equation for buses moving in the SAME direction (A to B).
The relative velocity of these buses with respect to the cyclist is $(V - 20)$.
The cyclist sees a bus cross him every $18\text{ mins}$ ($\frac{18}{60}\text{ hours}$). The relative distance they close in this time is the gap $VT$.

$$\frac{VT}{V - 20} = \frac{18}{60} \quad \text{--- (Equation 1)}$$

Step 3: Formulate equation for buses moving in the OPPOSITE direction (B to A).
The relative velocity of these buses with respect to the cyclist is $(V + 20)$.
The cyclist sees a bus cross him every $6\text{ mins}$ ($\frac{6}{60}\text{ hours}$).

$$\frac{VT}{V + 20} = \frac{6}{60} \quad \text{--- (Equation 2)}$$

Step 4: Solve for $V$ by dividing Equation 1 by Equation 2.
The $VT$ and the $60$ denominators will cancel out nicely.

$$\frac{\frac{VT}{V - 20}}{\frac{VT}{V + 20}} = \frac{18}{6}$$

$$\frac{V + 20}{V - 20} = 3$$

Cross-multiply:

$$V + 20 = 3(V - 20)$$

$$V + 20 = 3V - 60$$

$$2V = 80 \Rightarrow V = 40$$

$$\text{Speed of buses } (V) = 40\text{ km/h}$$

Step 5: Solve for the period $T$.
Substitute $V = 40$ into Equation 2 (or Eq 1). Let's use Eq 2.

$$\frac{40T}{40 + 20} = \frac{6}{60}$$

$$\frac{40T}{60} = \frac{1}{10}$$

$$40T = 6 \Rightarrow T = \frac{6}{40} \text{ hours}$$

Convert $T$ into minutes by multiplying by 60:

$$T = \frac{6}{40} \times 60 = \frac{360}{40} = 9$$

$$\text{Time period } (T) = 9\text{ minutes}$$