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?

Question

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 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/s<sup>2</sup>.At the instant the particle achieves its maximum positive x coordinate,how far is it from the origin? A) 36 m B) 20 m C) 45 m D) 27 m E) 37 m

+ 4) 0

) m/s².At the instant the particle achieves its maximum positive x coordinate,how far is it from the origin?

Quizplus · Accepted Answer

We can use the kinematic equations to solve for the position of the particle at the time it achieves its maximum positive x-coordinate.

From the given information:

Initial velocity, u = 6.0 m/s 
Acceleration, a = (-2.0, 4.0) m/s^2

Let the time at which the particle achieves its maximum positive x-coordinate be t_max.

Using the equation: x = ut + (1/2)at^2

At t = t_max, x will be maximum, and the velocity in the x-direction will be zero. Therefore:

0 = u_x + a_x t_max

t_max = u_x/a_x = 6.0/2.0 = 3 seconds

Substituting this value of t_max into the equation for x, we get:

x = (6.0)(3) + (1/2)(-2.0)(3)^2 = 9.0 m

Therefore, the particle is 9.0 m from the origin when it achieves its maximum positive x-coordinate. Option B is the closest to this value, and is therefore the correct answer.

A Particle Starts from the Origin at T = 0