The relation between the position x and time | vedclass

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(D) Given the relation: $t = \sqrt{x} + 3$.Rearranging for $x$: $\sqrt{x} = t - 3$.Squaring both sides: $x = (t - 3)^2$.Velocity $v$ is the derivative

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$-9$

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(D) Given the relation: $t = \sqrt{x} + 3$.Rearranging for $x$: $\sqrt{x} = t - 3$.Squaring both sides: $x = (t - 3)^2$.Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t - 3)^2 = 2(t - 3)$.When the velocity is zero: $2(t - 3) = 0$,which gives $t = 3 \ s$.Initial position at $t = 0 \ s$: $x_i = (0 - 3)^2 = 9 \ m$.Final position at $t = 3 \ s$: $x_f = (3 - 3)^2 = 0 \ m$.Displacement is defined as the change in position: $\Delta x = x_f - x_i = 0 - 9 = -9 \ m$.

The relation between the position $x$ and time $t$ of a particle moving in one dimension is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. When the velocity of the particle becomes zero,the displacement of the particle is ........ $m$.

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