The position x of a particle moving in one di | vedclass

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Question

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

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

(D) Given equation: $t = \sqrt{x} + 3$
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 velocity is zero: $2(t - 3) = 0 \implies t = 3 	ext{ s}$
Position at $t = 3 	ext{ s}$: $x = (3 - 3)^2 = 0 	ext{ m}$
Initial position at $t = 0 	ext{ s}$: $x_0 = (0 - 3)^2 = 9 	ext{ m}$
Displacement $\Delta x = x - x_0 = 0 - 9 = -9 	ext{ m}$
The magnitude of displacement is $9 	ext{ m}$.

The position $x$ of a particle moving in one dimension under the influence of a constant force is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. Find the displacement of the particle in $m$ when its velocity becomes zero.

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