The motion of a particle is defined by the relation x = t3 – 12t2 +36t +32, where x and t are expressed in feet and seconds, respectively. determine the time, position, and acceleration of the particle when v = 0 ft/s.

1) Time: 2 s, 6 s

The position of the particle is given by:

where t is the time in seconds and x is the position in feet.

The velocity of the particle can be found by differentiating the position:

and it is expressed in ft/s.

In order to find the time at which the velocity is v=0 ft/s, we substitute v=0 into the previous equation:

So the two solutions are

t = 2 s

t = 6 s

2) Position: x = 64 ft and x = 32 ft

The position at which the velocity of the particle is v = 0 can be found by susbtituting t = 2 and t = 6 into the equation for the position.

For t = 2 s, we have

For t = 6 s, we have

So the two positions are

x = 64 ft

x = 32 ft

3) Acceleration: and

The acceleration of the particle can be found by differentiating the velocity. We find:

And substituting t = 2 and t = 6, we find the acceleration when the velocity of the particle is zero:

So the two accelerations are

i think the answer is a, hope this is right, that would be my best guess.

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.