Exams › JEE Advanced › Physics
Two boats A and B move in a river, both having the same speed relative to the flowing river water. Boat A appears to move perpendicular to the river bank when viewed by an observer who is drifting along with the river current. Boat B actually moves perpendicular to the river bank as seen by a stationary observer on the ground. Which of the following statements are INCORRECT?
- As seen from the ground, boat B moves faster than boat A
- As seen from the ground, boat A moves faster than boat B
- As seen by the drifting observer (moving with current), boat B moves faster than boat A
- As seen by the drifting observer (moving with current), boat A moves faster than boat B
Correct answer: As seen from the ground, boat A moves faster than boat B
Solution
Boat A moves at speed u as seen from ground (since the drifting observer sees it go straight across, its ground velocity has no downstream component). Boat B's ground speed = sqrt(u² + v²) > u. So from the ground, B is faster than A (option A is correct, option B is incorrect).
Related JEE Advanced Physics questions
- A person riding in a hot-air balloon experiences an upward acceleration of 4.9 m/s². Exactly 2 seconds after the balloon is released from ground level, the person drops a small object. Taking g = 9.8 m/s², find the maximum height above the ground that the dropped object reaches.
- An object is released from rest at some height above the ground. In addition to gravity, air exerts a constant horizontal acceleration on the object throughout its fall. The path of the object as seen from the ground frame will be a
- A swimmer can cross a still river of width d to a point exactly opposite the starting bank in time t0. When the river has a current, the swimmer still manages to reach the directly opposite point (same path), but now takes time t1 > t0. What is the speed of the river current?
- A car can produce a maximum acceleration of 5 m/s² and its brakes can produce a maximum retardation of 10 m/s². Assuming the car starts from rest and must come to rest at a destination 1500 m away, find the minimum time (in seconds) to travel this distance.
- From the top of a 100 m tower, a particle is thrown vertically upward with speed 50 m/s at t = 0. Two seconds later, a second particle is projected vertically upward from the ground with the same speed 50 m/s. At what time after t = 0 are the two particles at the same height?
- The velocity of a particle traveling in a straight line is given by v(t) = 5 - 6*e^(-t/2) m/s, where time t is in seconds and t >= 0. At t = 0, the particle is at position x = 7 m. The position as a function of time is x(t) = k*t + l*e^(-t/2) + m. Find the numerical value of (k + m) / l.
⚔️ Practice JEE Advanced Physics free + battle 1v1 →