A particle starts from the origin at t = 0 with a velocity of 6.0 m/s and moves in the xy plane with a constant acceleration of (-2.0 + 4.0) m/s2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin

9sqrt4

Explanation:

Assuming you mean 6.0i m/s and (-2.0i + 4.0j) m/s^2:

The maximum positive x coordinate is when the time is 3 seconds, because the horizontal velocity will be zero, and therefore will be a turning point, where after that point the x coordinate will only be more negative.

After 3 seconds the x coordinate is 6*3+1/2*-2*3^2=9, and the y coordinate will be 1/2*4*3^2=18.

The distance from the origin is sqrt(9^2+18^2) = 9*sqrt(1+4)

object 2 because all the other ones are wrong

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~