Set 8k

Q1 [3 Marks]: Write the mathematical expression for the relative displacement of an object A with respect to object B, if they start from different initial positions.

Step 1: Write the position equations for both objects.
Let object A start from initial position $x_{A0}$ with uniform velocity $v_A$. Its position at time $t$ is:

$$x_A(t) = x_{A0} + v_A t$$

Let object B start from initial position $x_{B0}$ with uniform velocity $v_B$. Its position at time $t$ is:

$$x_B(t) = x_{B0} + v_B t$$

Step 2: Find the relative displacement.
The relative position (or relative displacement) of A with respect to B at any time $t$ is the difference between their individual positions:

$$x_{AB}(t) = x_A(t) - x_B(t)$$

$$x_{AB}(t) = (x_{A0} + v_A t) - (x_{B0} + v_B t)$$

Grouping the initial positions and the velocity terms together gives:

$$x_{AB}(t) = (x_{A0} - x_{B0}) + (v_A - v_B)t$$

Q2 [3 Marks]: A jet airplane traveling at the speed of $500\text{ km/h}$ ejects its products of combustion at the speed of $1500\text{ km/h}$ relative to the jet plane. What is the speed of the combustion products with respect to an observer on the ground?

Step 1: Establish a sign convention.
Let the forward direction of the jet be positive (+ve).
Velocity of the jet relative to the ground, $v_{jet} = +500\text{ km/h}$.

Step 2: Identify the relative velocity.
The combustion products are ejected out the back of the engine, meaning they move in the opposite direction of the jet. Therefore, their velocity relative to the jet ($v_{gas/jet}$) is negative.

$$v_{gas/jet} = -1500\text{ km/h}$$

Step 3: Apply the relative velocity formula.
By definition, $v_{gas/jet} = v_{gas} - v_{jet}$ (where all single-subscript velocities are with respect to the ground).

$$-1500 = v_{gas} - 500$$

$$v_{gas} = -1500 + 500 = -1000\text{ km/h}$$

The negative sign indicates the gases are moving backward relative to a stationary observer on the ground. Since the question asks for speed (magnitude):

$$\text{Speed of combustion products } = 1000\text{ km/h}$$

Q3 [4 Marks]: Two cars A and B are running at velocities of $60\text{ km/h}$ and $45\text{ km/h}$. Calculate the relative velocity of car A with respect to car B if they are moving (i) in the same direction, and (ii) in opposite directions.

Case (i): Moving in the SAME direction
Let both cars be moving in the positive direction.
$v_A = +60\text{ km/h}$, $v_B = +45\text{ km/h}$.

$$v_{AB} = v_A - v_B$$

$$v_{AB} = 60 - 45$$

$$v_{AB} = 15\text{ km/h}$$

Case (ii): Moving in OPPOSITE directions
Let car A move in the positive direction and car B move in the negative direction.
$v_A = +60\text{ km/h}$, $v_B = -45\text{ km/h}$.

$$v_{AB} = v_A - v_B$$

$$v_{AB} = 60 - (-45)$$

$$v_{AB} = 60 + 45$$

$$v_{AB} = 105\text{ km/h}$$

Q4 [5 Marks]: A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of $72\text{ km/h}$. The jeep follows it at a speed of $90\text{ km/h}$, crossing the turning ten seconds later than the bike. Assuming they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Step 1: Convert speeds to m/s.
Velocity of motorbike, $v_b = 72 \times \frac{5}{18} = 20\text{ m/s}$.
Velocity of jeep, $v_j = 90 \times \frac{5}{18} = 25\text{ m/s}$.

Step 2: Calculate the "head start" of the motorbike.
The bike crosses the turning and travels for 10 seconds before the jeep even reaches the turning.

$$\text{Distance covered by bike in 10s } (d) = 20\text{ m/s} \times 10\text{ s} = 200\text{ m}$$

So, when the jeep crosses the turning, the bike is exactly $200\text{ m}$ ahead.

Step 3: Apply Relative Velocity.
Now, treat the jeep as the observer. The relative velocity of the jeep closing in on the bike is:

$$v_{rel} = v_j - v_b = 25 - 20 = 5\text{ m/s}$$

This means the jeep closes the gap by $5\text{ m}$ every second.

Step 4: Find the time taken to catch up.
The total relative distance to close is the $200\text{ m}$ head start.

$$\text{Time to catch up } (t) = \frac{\text{Relative Distance}}{\text{Relative Velocity}} = \frac{200}{5} = 40\text{ s}$$

Note: This is 40 seconds AFTER the jeep crosses the turning.

Step 5: Calculate the distance from the turning.
The jeep travels for $40\text{ s}$ at its speed of $25\text{ m/s}$ from the turning.

$$\text{Total Distance} = v_j \times t = 25 \times 40$$

$$\text{Distance from turning } = 1000\text{ m (or 1 km)}$$

Position-Time (x-t) Catch-up Graph

Notice how the Jeep's line is steeper (higher velocity). It starts 10 seconds late but eventually intersects the bike's line exactly at $1000\text{ m}$.