River-Boat Problems — Crossing shortest path vs shortest time
Kinematics
13
JEE Qs
8%
Hard
60
min
Master vector resolution and relative velocity principles; visualize the velocity vectors in the ground frame for both scenarios (shortest time vs. shortest path) to avoid common pitfalls.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always resolve velocities into components parallel and perpendicular to the river flow (or current direction) when analyzing motion.
- 2The time to cross the river depends solely on the component of the boat's velocity (relative to ground) perpendicular to the river flow.
- 3The drift (displacement along the river flow) depends on the component of the boat's velocity (relative to ground) parallel to the river flow and the time taken to cross.
- 4For shortest time, the boat's velocity relative to the river (v_br) should be directed perpendicular to the river flow.
- 5For shortest path (zero drift), the boat's resultant velocity relative to the ground (v_b) must be directed perpendicular to the river flow. This requires the boat to head upstream at a specific angle, which is only possible if the boat's speed relative to the river (v_br) is greater than or equal to the river's speed (v_r).
⚠️ Common Mistakes
- ✕Confusing the boat's velocity relative to the river (v_br) with its velocity relative to the ground (v_b). Always use v_b = v_br + v_r for analysis in the ground frame.
- ✕Incorrectly identifying which velocity component (parallel or perpendicular to flow) determines crossing time versus drift.
- ✕Failing to check the condition v_br >= v_r when attempting to achieve a 'shortest path' (zero drift) crossing.
📝 Practice Questions
See allQ27.A ball having kinetic energy KE, is projected at an angle of 60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ? (1) (KE) (2) (KE) 8 2 (3) (KE) (4) (KE) 16 4
Q28.The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4 s ? (1) 30 m (2) 11 m (3) 10 m (4) 13 m
Q50.A particle is projected at an angle of 30∘ from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1 . The ratio h0 : h1 is _________ [Take, g = 10 m/s2 ]
Q43.The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km . (1) 12 (2) 3 (3) 6 (4) 9
Q27.The position vector of a moving body at any instant of time is given as →r = . The magnitude (5t2^i −5t^j)m and direction of velocity at t = 2 s is, (1) 5√15 m/s, making an angle of tan−1 4 with - ve (2) 5√15 m/s, making an angle of tan−1 4 with + ve Y axis X axis (3) 5√17 m/s, making an angle of tan−1 4 with + ve (4) 5√17 m/s, making an angle of tan−1 4 with - ve X axis Y axis
Q31.Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α) and (45∘+ α), respectively, with the horizontal direction. The ratio of their maximum heights attained is : (1) 1−tan α (2) 1−sin 2α 1+tan α 1+sin 2α (3) 1+sin 2α (4) 1+sin α 1−sin 2α 1−sin α
NCERT Chapters
- Class 11 Physics Ch 4: Motion in a Plane