Pls answer the question

Dear Student,

t = \sqrt{x} + 5 \sqrt{x} = t - 5 x = {(t - 5)}^{2} = t^{2} + 25 - 10 t v = \frac{dx}{dt} = 2 t - 10 The time at which velocity of particle is zero, 2 t - 10 = 0 t = 5 \sec . Now, initial position of particle, x_{0} = {(0)}^{2} + 25 - 10 (0) = 25 m Position of particle at t = 5 is, x = {(5)}^{2} + 25 - 10 (5) = 0 displacement = x - x_{0} = 0 - 25 = - 25 m

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