A particle leaves the origin with an initial velocity v Overscript right-arrow EndScripts equals left-parenthesis 7.34 i Overscript ̂ EndScripts right-parenthesis m divided by s and a constant acceleration a Overscript right-arrow EndScripts equals left-parenthesis negative 2.44 i Overscript ̂ EndScripts minus 1.04 j Overscript ̂ EndScripts right-parenthesis m divided by s Superscript 2. When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector?

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To find the velocity and position vector of the particle when it reaches its maximum x-coordinate, we need to first find the time it takes for the particle to reach this point.

The particle starts at the origin and moves under constant acceleration. We can use the kinematic equation to find the time it takes for the particle to reach its maximum x-coordinate.

The kinematic equation relating displacement (x), initial velocity (v), time (t), and acceleration (a) is:

x = v*t + (1/2)*a*t^2

Since we want to find the time when the particle reaches its maximum x-coordinate, we can set the displacement (x) equal to this maximum value. Let's call this maximum x-coordinate "xmax."

xmax = v*tmax + (1/2)*a*tmax^2

Now, let's plug in the given values:

v = 7.34 i m/s
a = (-2.44 i - 1.04 j) m/s^2

We need to find tmax, so we will rearrange the equation to isolate tmax:

xmax = v*tmax + (1/2)*a*tmax^2
0 = (1/2)*a*tmax^2 + v*tmax - xmax

Now, we can use the quadratic formula to solve for tmax:

tmax = (-b ± √(b^2 - 4ac)) / (2a)

where:
a = (1/2)*a = (1/2)*(-2.44 i - 1.04 j) = (-1.22 i - 0.52 j) m/s^2
b = v = 7.34 i m/s
c = -xmax

Plugging in these values:

tmax = [-(7.34 i) ± √((7.34 i)^2 - 4*(-1.22 i - 0.52 j)*(-xmax))] / (2*(-1.22 i - 0.52 j))

Simplifying further:

tmax = [-(7.34 i) ± √(53.7956 i^2 - 4*(-1.22 i - 0.52 j)*(-xmax))] / (-2.44 i + 1.04 j)
tmax = [-(7.34 i) ± √(-53.7956 + 4*(-1.22 i - 0.52 j)*(-xmax))] / (-2.44 i + 1.04 j)

Since the expression inside the square root is negative, this means the motion is constrained in the y-axis direction. Therefore, the maximum x-coordinate is reached when the y-coordinate is stationary.

To solve for tmax, we can set the y-component of acceleration to zero:

-1.04 j = 0

This equation yields j = 0, meaning there is no y-component acceleration. Thus, the maximum x-coordinate is reached when the y-coordinate is stationary.

Now that we know the maximum x-coordinate, we can use the kinematic equation to find the time it takes for the particle to reach this point:

xmax = v*tmax

Solving for tmax:

tmax = xmax / v

After finding tmax, we can find the velocity and position vector of the particle at that time.

(a) velocity:
To find the velocity, we can use the equation:

v = v0 + a*t

Given:
v0 = v (initial velocity)
a = a (acceleration)
t = tmax

v = v0 + a*t
v = v + a*tmax

Substituting the given values, we can calculate the velocity.

(b) position vector:
To find the position vector at time tmax, we can use the equation:

r = r0 + v0*t + (1/2)*a*t^2

Given:
r0 = 0 (initial position vector)
t = tmax
v0 = v (initial velocity)
a = a (acceleration)

Substituting the given values, we can calculate the position vector.

Please provide the value of xmax to proceed with the calculations.