The position of a particle moving along x-axis varies with time t as $4 x + t^{2} = 4$ $4x + t^2 = 4$ . At what position, the particle is at rest?

Question

x = 4

x = -2

x = 1

x = 2

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Answer 1 · 2017-11-20T02:49:39

Selection 3. x = 1

Given:

$4 x + t^{2} = 4$ $4x + t^2 = 4$

Solve for x:

$4 x = 4 - t^{2}$ $4x = 4 - t^2$

$x = 1 - \frac{t^{2}}{4} [1]$ $x = 1 - t^2/4" [1]"$

Compute the first derivative with respect to t:

$\frac{d x}{d t} = \frac{t}{2}$ $dx/dt = t/2$

The speed is 0 at the time(s) where the first derivative is equal to 0:

$\frac{t}{2} = 0$ $t/2 = 0$

$t = 0$ $t = 0$

To find the x coordinate, substitute 0 for t in equation [1] :

$x = 1 - \frac{0^{2}}{4}$ $x = 1 - 0^2/4$

$x = 1$ $x = 1$

This corresponds to selection 3.