Two cars, A and B, start from rest at the same point. Car A moves with constant acceleration 'a' and car B moves with constant acceleration '2a'. After how much time will the distance between them be 'S'?

A body executing simple harmonic motion has a maximum acceleration equal to $24\,\,m{s^{ - 2}}$ and maximum velocity of $16\,\,m{s^{ - 1}}$ ,the amplitude of the simple harmonic motion is

The equation of SHM is given by $x = 3\sin 20\pi t + 4\cos 20\pi t$ Where x is in cm and t is I second. The amplitude is

The amplitude of a particle performing SHM is ‘a’. The displacement at which its velocity will be a half the maximum velocity is

A particle of 0.25 kg vibrates with a period of 2 second. If its greatest displacement is 0.4 m the maximum velocity in $m{s^{ - 1}}$ is

If the maximum acceleration of a $SHM$ is $\alpha$ and the maximum velocity is $\beta$, then the amplitude of vibration is

For a body in SHM the velocity is given by the relation $V = \sqrt {144 - 16{x^2}} \,\,m{s^{ - 1}}.$ The maximum acceleration is

In SHM, the amplitude is $6$ cm. At what displacement from the mean position will the magnitude of velocity be equal to half the maximum velocity, if at a certain displacement, the magnitudes of velocity and acceleration are equal?

The equation of SHM is given by $x = 3\sin 20\pi t + 4\cos 20\pi t$ Where x is in cm and t is I second. The amplitude is

A body executing simple harmonic motion has a maximum acceleration equal to $24\,\,m{s^{ - 2}}$ and maximum velocity of $16\,\,m{s^{ - 1}}$ ,the amplitude of the simple harmonic motion is

The maximum velocity of a particle executing SHM is $V$ . If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value, the maximum velocity becomes