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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

Respuesta :

tqiu

Answer:

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)

onyebuchinnaji

At the instant the particle achieves its maximum positive x coordinate, the displacement is 4.03 m.

Displacement of the particle

The displacement of the particle is calculated as follows;

v² = u² + 2as

where;

  • a is acceleration

a = √(2² + 4²) = 4.47 m/s²

0 = u² + 2as

s = -u²/2a

s = -(6²)/(2 x 4.47)

s = -4.03 m

Thus, at the instant the particle achieves its maximum positive x coordinate, the displacement is 4.03 m.

Learn more about displacement here: https://brainly.com/question/321442

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