A particle travels along a path described by the function y=0.2x^3. The exponent of velocity is given by vx=(5t^2)ft/s. When t=0, x=y=0, find the particle's distance from the origin and the magnitude of its acceleration when t=1s.

To find the distance from the origin, we need to find the position of the particle at t=1s. The position can be determined by substituting the value of t into the equation y=0.2x^3.

Given: t = 1s
Substitute t = 1 into the equation:
y = 0.2x^3

To find the value of x, we need additional information. The function y=0.2x^3 does not provide enough information to determine the value of x.

However, we can find the magnitude of the particle's acceleration when t = 1s by finding the second derivative of the position function y=0.2x^3 with respect to time.

Given: y = 0.2x^3
Differentiate both sides of the equation with respect to time:
d^2y/dt^2 = d^2(0.2x^3)/dt^2

To find the second derivative, we need to differentiate the equation again with respect to x.

d^2(0.2x^3)/dt^2 = 0.2 * d^2(x^3)/dt^2

Differentiate x^3 with respect to x:
d^2(x^3)/dt^2 = d^2(3x^2)/dt^2

Differentiate 3x^2 with respect to x:
d^2(3x^2)/dt^2 = 3 * d^2(x^2)/dt^2

Differentiate x^2 with respect to x:
d^2(x^2)/dt^2 = d^2(2x)/dt^2

Differentiate 2x with respect to x:
d^2(2x)/dt^2 = 2 * d^2(x)/dt^2

Differentiate x with respect to x:
d^2(x)/dt^2 = 0

Now, substitute the results back into the equation:

d^2y/dt^2 = 0.2 * 3 * 3 * 2 * 2 * 0 = 0

The magnitude of the particle's acceleration when t = 1s is 0 ft/s^2.