Exams › JEE Main › Physics

A particle is moving with velocity v = k(yî + xĵ), where k is a constant. The general equation for its path is

1. y = x² + constant
2. y² = x + constant
3. xy = constant
4. y² = x² + constant

Correct answer: y² = x² + constant

Solution

vx = dx/dt = k y and vy = dy/dt = k x, so dy/dx = x/y -> y dy = x dx -> y^2/2 = x^2/2 + C -> y^2 = x^2 + constant.

Related JEE Main Physics questions

• A person travels 30 m due north, then 20 m due east, and finally 30√2 m in the south-west direction. What is the person's net displacement from the starting point?
• Two particles are set in motion at the same instant from a common point along two different straight paths. One particle moves with constant speed v, while the other starts with uniform acceleration a. If the angle between their directions of motion is α, then the time at which the relative velocity becomes minimum is
• Two time-dependent vectors are given by A = cos(ω t) î + sin(ω t) ĵ and B = cos ((ω t)/(2)) î + sin ((ω t)/(2)) ĵ. At what value of t are these vectors perpendicular to each other?
• A bus travels north along a straight road at a constant speed of 50 km/h and then takes a 90° turn. If its speed does not change after the turn, the change in the bus’s velocity during the turn is
• Two vectors A and B combine to give a resultant that is perpendicular to A, and the magnitude of this resultant is one-half the magnitude of B. What is the angle between A and B?
• A person moves on a straight path with constant velocity u î. He observes that rain appears to fall straight downward along -ĵ. When he increases his speed to twice the original value, the rain is seen to make an angle θ with the vertical. The velocity of the rain relative to the ground is

⚔️ Practice JEE Main Physics free + battle 1v1 →