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Motion in a Plane

JEE Main 2024 (Online) 29th January Evening ShiftNumericalMedium📈 Topic in 7% papers+4
Ques. 1JEE Main 2024, 29 January (Evening Shift)

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is 4 m/s, the time taken to complete the first revolution will be 1 [1-e-2 ]s, where = .

Answer

8

Solution

at=an v dvds= v2R

dvv= dsR v=v0 es/R

dt= dsv t= 1v0 02 Re-s/R ds

= Rv0 (1-e-2 )

= 0.54 (1-e-2 )= 18 (1-e-2 ) s

=8

Answer: 8

Keep practising — Motion in a Plane

Ques. 2JEE Main 2024, 27 January (Morning Shift)

Position of an ant (S in metres) moving in Y-Z plane is given by S=2t2 j+5 k (where t is in second). The magnitude and direction of velocity of the ant at t=1 s will be :

  1. 16 m/s in y-direction
  2. 4 m/s in x-direction
  3. 9 m/s in z-direction
  4. 4 m/s in y-direction
Ques. 3JEE Main 2024, 27 January (Morning Shift)

A particle starts from origin at t = 0 with a velocity 5 i m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3 i + 2 j) m/s2. If the x-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is m/s. The value of is .

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