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Saturday, 31 October

JEE Main 2026 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2026, Shift 2, Physics Q45

Question 45 of 75 in this shift, Physics question 20 of 25, Section A.

The equation of motion of a particle is given by x=asin⁡(50t+π3)x = a\sin\left(50t + \frac{\pi}{3}\right) cm. The particle will come to rest at time t1t_1 and it will have zero acceleration at time t2t_2. The t1t_1 and t2t_2 respectively are
  1. (1)π300\frac{\pi}{300} s, π75\frac{\pi}{75} sOfficial answer
  2. (2)π75\frac{\pi}{75} s, π300\frac{\pi}{300} s
  3. (3)π300\frac{\pi}{300} s, π25\frac{\pi}{25} s
  4. (4)π50\frac{\pi}{50} s, π100\frac{\pi}{100} s

Official answer

Option 1

NTA final key.