Q70 / 75 · JEE Main 2026 (02 April Shift 2) · Physics

The equation of motion of a particle is given by $x = a \sin\left(50t + \dfrac{\pi}{3}\right)$ cm. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$. The $t_1$ and $t_2$ respectively are _______.

  • Correct answer
    A
    $\dfrac{\pi}{300} \, s, \dfrac{\pi}{75} \, s$
  • B
    $\dfrac{\pi}{75} \, s, \dfrac{\pi}{300} \, s$
  • C
    $\dfrac{\pi}{300} \, s, \dfrac{\pi}{25} \, s$
  • D
    $\dfrac{\pi}{50} \, s, \dfrac{\pi}{100} \, s$

Solution

The equation of motion is given by $x = a \sin\left(50t + \dfrac{\pi}{3}\right)$.

Velocity is the rate of change of displacement:
$v = \dfrac{dx}{dt} = 50a \cos\left(50t + \dfrac{\pi}{3}\right)$

The particle comes to rest when $v = 0$. For the first time it comes to rest:
$\cos\left(50t_1 + \dfrac{\pi}{3}\right) = 0$

$50t_1 + \dfrac{\pi}{3} = \dfrac{\pi}{2}$

$50t_1 = \dfrac{\pi}{2} - \dfrac{\pi}{3} = \dfrac{\pi}{6}$

$t_1 = \dfrac{\pi}{300} \text{ s}$

Acceleration is the rate of change of velocity:
$A = \dfrac{dv}{dt} = -2500a \sin\left(50t + \dfrac{\pi}{3}\right)$

The particle has zero acceleration when $A = 0$. For the first time it has zero acceleration:
$\sin\left(50t_2 + \dfrac{\pi}{3}\right) = 0$

$50t_2 + \dfrac{\pi}{3} = \pi$

$50t_2 = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$

$t_2 = \dfrac{2\pi}{150} = \dfrac{\pi}{75} \text{ s}$

Answer: $\dfrac{\pi}{300} \, s, \dfrac{\pi}{75} \, s$

Hence, the correct answer is option: 'A'.

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