# The particle travels along the path defined by the parabola y=0.5x^2. If t

The particle travels along the path defined by the parabola $y=0.5{x}^{2}$. If the component of velocity along the x axis is vx = (5t) ft/s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.
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David Tyson
From the equation for ${v}_{x}$ we can get an equation for x:
$x=2.5{t}^{2}$
We can insert that to find the equation for y as a function of time:
$y=0.5\cdot {2.5}^{2}\cdot {t}^{4}=3.125{t}^{4}$
Distance at the t=1 is:
${d}_{x}=2.5$ ft
${d}_{y}=3.125ft$
$d=\sqrt{{d}_{x}+{d}_{y}}=\sqrt{{2.5}^{2}+{3.125}^{2}}$
$d=4$ ft
Acceleration in t=1 is:

$a=\sqrt{{a}_{x}+{a}_{y}}=\sqrt{{5}^{2}+{37.5}^{2}}$

Result: