Problem Statement
Solve the kinematics problem: Train A moves at 60 km/h and train B moves at 40 km/h in the same direction. Find the velocity of A relative to B. Relative velocity = $v_A – v_B$ (same direction) Step 1: $v_{AB} = v_A – v_B = 60 – 40 = 20$ km/h in the direction of motion. $$\boxed{v_{AB} = 20\text{ km/h (in direction of motion)}}$
Given Information
- $\boxed{v_{AB} = 20\text{ km/h (in direction of motion)}$
Physical Concepts & Formulas
Relative velocity $\mathbf{v}_{AB} = \mathbf{v}_A – \mathbf{v}_B$ is the velocity of A as seen from B’s reference frame. For river-crossing problems, the boat’s velocity relative to ground = boat velocity relative to water + water velocity. The angle for minimum crossing time differs from the angle for minimum drift.
- $\mathbf{v}_{A/B} = \mathbf{v}_A – \mathbf{v}_B$
- River crossing: $\mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{boat/water}} + \mathbf{v}_{\text{river}}$
- Minimum time: boat aims perpendicular to river bank
- Minimum drift: $\sin\phi = v_{\text{boat}}/v_{\text{river}}$ (if river faster)
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{v_{AB} = 20\text{ km/h (in direction of motion)}}$
Given Information
- Initial velocity $u$ (or $v_0$)
- Acceleration $a$ (constant unless stated otherwise)
- Time $t$ or distance $s$ as given
Physical Concepts & Formulas
Kinematics describes motion without reference to its cause. For constant acceleration, the four SUVAT equations are sufficient to solve any problem. They follow directly from the definitions of velocity ($v = ds/dt$) and acceleration ($a = dv/dt$). For 2D problems (projectile motion), the horizontal and vertical motions are independent — horizontal: constant velocity; vertical: constant acceleration $g$ downward. Relative motion problems require defining a reference frame explicitly and using vector subtraction.
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- $s = \tfrac{1}{2}(u+v)t$
- Range of projectile: $R = \dfrac{u^2\sin 2\theta}{g}$
- Max height: $H = \dfrac{u^2\sin^2\theta}{2g}$
Step-by-Step Solution
Step 1 — List knowns and unknown: $u$, $v$, $a$, $s$, $t$ — identify which three are known.
Step 2 — Choose the SUVAT equation that contains the unknown and all three known quantities.
Step 3 — Substitute and solve algebraically.
Step 4 — For 2D: Decompose $\vec{u}$ into $u_x = u\cos\theta$, $u_y = u\sin\theta$. Solve $x$ and $y$ separately.
Worked Calculation
Substituting all values with units:
Projectile at $u = 20\,\text{m/s}$, $\theta = 30°$:
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Answer
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Answer
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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