Problem Analysis: Motion in Opposite Directions

This problem involves two individuals, A and B, moving in opposite directions from a common starting point. We are given their relative speeds and a specific scenario where A travels a certain distance, turns back, and we need to find where A crosses B relative to the start.

Calculating Speeds

We are given:

• Speed of B: $v_B = 1$ km/h.
• Speed of A is twice the speed of B: $v_A = 2 \times v_B = 2 \times 1 \text{ km/h} = 2$ km/h.

Determining Positions When A Turns Back

A travels 2 km before turning back. The time taken for A to cover this distance is:

$ t_1 = \frac{\text{Distance}}{\text{Speed of A}} = \frac{2 \text{ km}}{2 \text{ km/h}} = 1 \text{ hour} $

During this 1 hour, B also travels in the opposite direction. The distance B covers is:

$ d_B = v_B \times t_1 = 1 \text{ km/h} \times 1 \text{ h} = 1 \text{ km} $

Let the starting point be the origin (0 km). If A moves in the positive direction, A is at $+2$ km and B is at $-1$ km when A turns back.

Finding the Crossing Point

A turns back, so A's velocity relative to the origin becomes $-2$ km/h. B continues in the negative direction with velocity $-1$ km/h.

Let $t$ be the time in hours after A turns back.

• A's position at time $t$: $x_A(t) = (\text{Position when turned back}) + (\text{Velocity of A}) \times t$ $ x_A(t) = 2 + (-2)t = 2 - 2t $
• B's position at time $t$: $x_B(t) = (\text{Position when A turned back}) + (\text{Velocity of B}) \times t$ $ x_B(t) = -1 + (-1)t = -1 - t $

A crosses B when their positions are equal:

$ x_A(t) = x_B(t) $

$ 2 - 2t = -1 - t $

Solving for $t$:

$ 2 + 1 = 2t - t $

$ 3 = t $

So, they cross 3 hours after A turns back.

Calculating Distance from Starting Point

We can find the position where they cross using either A's or B's position equation at $t=3$ hours.

Using A's position:

$ x_A(3) = 2 - 2(3) = 2 - 6 = -4 \text{ km} $

The distance from the starting point is the absolute value of the position:

$ \text{Distance} = |-4 \text{ km}| = 4 \text{ km} $