Physics · Motion In A Straight Line · NEET
Always use one master formula: v(A relative to B) = v(A) - v(B). You never truly "add"; you always subtract, but you keep signs. Pick a positive direction first (say, up or right). If A and B move the same way, both velocities are positive, so v(A) - v(B) is a small subtraction. If they move opposite ways, one velocity is negative, so subtracting a negative becomes an addition and the relative speed is large. Example: two buses at 40 and 30 m/s. Same direction: 40 - 30 = 10 m/s. Opposite direction: 40 - (-30) = 70 m/s. This is why an oncoming bus flashes past you but a same-direction bus seems to crawl.
Because SPEEDS add, not times. When Preeti walks up a moving escalator, her speed relative to the ground is her walking speed PLUS the escalator's speed. Let the escalator length be L. Walking alone: speed = L/t1. Escalator alone: speed = L/t2. Together: L/t = L/t1 + L/t2. Cancel L: 1/t = 1/t1 + 1/t2. This gives t = t1*t2 / (t1 + t2). The trap is adding the times directly; that would make her SLOWER, which is nonsense because both effects help her go up.
Only if the lift moves with UNIFORM (constant) velocity. Then the lift is an inertial frame, there is no pseudo-force, and the coin falls with the same g over the same height, so the time is identical (t1 = t2). Whether the lift goes up or down does not matter, as long as the speed is constant. It changes ONLY if the lift ACCELERATES: then the effective gravity relative to the lift floor becomes g + a (lift accelerating up) or g - a (lift accelerating down), and the fall time changes.
A pseudo force is a fake force you add ONLY when you sit inside an accelerating (non-inertial) frame like a rising lift. Its size is m*a and it points OPPOSITE to the lift's acceleration. If the lift accelerates upward with a, an object inside feels effective gravity g_eff = g + a (heavier). If the lift accelerates downward with a, g_eff = g - a (lighter); if a = g (free fall) the object floats. For NEET, use pseudo force only when the frame speeds up or slows down, never for constant-velocity lifts.
The spacing between buses is fixed: d = V(bus) * T, where T is the launch interval. A same-direction bus overtakes you at relative speed (V_bus - V_you), so the gap it closes is d / (V_bus - V_you) = time gap. An opposite bus meets you at relative speed (V_bus + V_you), giving d / (V_bus + V_you). Write both equations, divide one by the other to cancel d and T, solve for V_bus, then back-substitute. This is exactly the NEET 2025 bus problem worked below.
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be:
Buses leave cities X and Y in both directions at regular intervals of T minutes with the same speed. A girl driving from X to Y at 60 km/h notices that a bus moving in her direction passes her every 30 min, while a bus moving in the opposite direction passes her every 10 min. The interval T and the speed of the buses are respectively:
A person in a lift drops a coin. The coin takes time t1 to reach the floor when the lift is at rest, and time t2 when the lift is moving up with uniform velocity. Then:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a vector. In one dimension it has a sign (+ or -) that tells direction. Its magnitude is the relative speed. Always fix a positive direction before plugging numbers into v(A rel B) = v(A) - v(B).
For an oncoming train the velocities are opposite, so relative speed = your speed + its speed (large). For a same-direction train they subtract, giving a small relative speed. Same physics as the bus problem.
No, not for uniform velocity. Constant velocity in any direction means zero acceleration, so the coin falls in the same time as in a lift at rest. Direction only matters when the lift accelerates.
When the lift accelerates up with a, effective gravity is g + a. When it accelerates down with a, it is g - a. When it accelerates down at a = g (free fall), effective gravity is zero and objects float. Use this for fall-time and apparent-weight questions.
It appears regularly: NEET 2017 (escalator), 2019 (lift coin) and 2025 (bus interval) all used it. The escalator reciprocal trick and the bus spacing method are high-yield, so memorise both patterns.
