The displacement of a particle moving in one dimension under the actio

1.	The displacement of a particle moving in one dimension under the action of a constant force is related to the time t be the equation t = \(\sqrt{x}\) + 3, where x is in metre and t is in second. Find the displacement of the particle when its velocity is zero.
Answer» \(t=\sqrt{x}+3\) or \(\sqrt{x}\) = t - 3 ...(i) Squaring on both sides of equations (i), = ( t – 3)² …(ii) If v be the velocity of the particle, Then \(v=\frac{dx}{dt}\) = \(\frac{d}{dt}(t^2-6t+9)\) = 2t - 6 when v = 0, 2t - 6 = 0 or t = 3 sec. ...(iii) ∴ From equations (ii) and (iii), x = 3² + 9 - 6 x 3 = 18 - 18 x = 0

The displacement of a particle moving in one dimension under the action of a constant force is related to the time t be the equation t = \(\sqrt{x}\) + 3, where x is in metre and t is in second. Find the displacement of the particle when its velocity is zero.

Answer»

\(t=\sqrt{x}+3\)

or \(\sqrt{x}\) = t - 3 ...(i)

Squaring on both sides of equations (i),

= ( t – 3)² …(ii)

If v be the velocity of the particle,

Then \(v=\frac{dx}{dt}\)

= \(\frac{d}{dt}(t^2-6t+9)\)

= 2t - 6

when v = 0, 2t - 6 = 0

or t = 3 sec. ...(iii)

∴ From equations (ii) and (iii),

x = 3² + 9 - 6 x 3

= 18 - 18

x = 0