The position of a particle moving along the $x$ axis varies in time according to the expression $x = 3 t^{2}$ , where $x$ is in meters and $t$ is in seconds. Evaluate its position (a) at $t = 3.00 s$ and (b) at $3.00 s + Δ t$ . (c) Evaluate the limit of $Δ x / Δ t$ as $Δ t$ approaches zero to find the velocity at $t = 3.00 s$ .

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Solution

a) At any time, $t$ , the position is given by $x = (3.00 m / s^{2}) t^{2}$ .
Thus, at $t_{i} = 3.00 s : x_{i} = (3.00 m / s^{2}) (3.00 s)^{2} = 27.0 m$ .

(b) At $t_{f} = 3.00 s + Δ t :: x_{f} = (3.00 m / s^{2}) (3.00 s + Δ t)^{2}$ , or
$x_{f} = 27.0 m + (18.0 m / s) Δ t + (3.00 m / s^{2}) (Δ t)^{2}$ .

(c) The instantaneous velocity at $t = 3.00 s$ is:
${lim}_{Δ t \to 0} \frac{Δ x}{Δ t} = {lim}_{Δ t \to 0} \frac{(18.0 m / s) Δ t + (3.00 m / s^{2}) (Δ t)^{2}}{Δ t}$
$= {lim}_{Δ t \to 0} (18.0 m / s) + (3.00 m / s^{2}) (Δ t) = 18.0 m / s$

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