Question

Relation between displacement x and time t for a moving particle under influence of a force, is t = \(\sqrt{x}\) + 3, where x is in meters and t is in second, find:

(i) Displacement of the particle at the moment when its velocity is zero.

(ii) Work done by the force in first 6 seconds.

Answer 1

\(\therefore\ t = \sqrt{x} + 3\ or\ \sqrt{x} = t-3\)

or x = (t - 3)² = t² - 6t + 9

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

= 2t - 6

(i) When v = 0,

then 2t - 6 = 0 ⇒ t = 3

∴ Displacement when v = 0

x = t² - 6t + 9 = (3)² - 6 x 3 + 9

= 9 - 18 + 9 = 18 - 18 = 0

or x = 0 (zero)

(ii) ∵ x = t² - 6t + 9

∴ At t = 0, x₀ = 0 = 0 - 0 + 9 = 9

at t = 6s, x₆ = (6)² - 6 x 6 + 9 = 36 - 36 + 9

or x₆ = 9 m

∴ Displacement ∆x = x₆ - x0 = 9 - 9 = 0

Therefore work done

W = F.∆x = F x 0 = 0

or W = 0