A particle leaves the origin with an initial velocity = (6.91 ) m/s and a constant acceleration

= ( - 2.38 - 2.79 ) m/s2. When the particle reaches its maximum x coordinate, what are
(a) its velocity, (b) its position vector?

I completely agree with Ms Sue. I respond to a fair share of the physics questions here, and am reluctant to help students who dump a lot of questions and show no work or evidence of thought at all.

Posting a lot of questions under different names, as some do, is unlikely to fool us.

I am happy to explain out how to do some problems, but you will have to do the work and show what you have learned.

To find the velocity and position vector of the particle when it reaches its maximum x coordinate, we can use the equations of motion.

(a) Velocity at the maximum x coordinate:
The velocity of the particle can be found using the equation v = u + at, where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given:
Initial velocity, u = 6.91 m/s
Acceleration, a = (-2.38 - 2.79) m/s^2 = -5.17 m/s^2 (combining the x and y components)

To find the time it takes to reach the maximum x coordinate, we need to find the time when the particle comes to rest momentarily. This occurs when the velocity becomes zero. So, we have:
0 = 6.91 + (-5.17)t
5.17t = 6.91
t = 6.91 / 5.17
t ≈ 1.335 seconds

Now that we have the time, we can find the velocity:
v = u + at
v = 6.91 + (-5.17)(1.335)
v ≈ -6.92 m/s

Therefore, when the particle reaches its maximum x coordinate, its velocity is approximately -6.92 m/s.

(b) Position vector at the maximum x coordinate:
To find the position vector, we need to know the displacement from the origin at the maximum x coordinate. The displacement can be found using the equation s = ut + (1/2)at^2, where:
s = displacement

Using the time t ≈ 1.335 seconds, we can calculate the displacement:
s = (6.91)(1.335) + (1/2)(-5.17)(1.335)^2
s ≈ 4.623 + (-4.337)
s ≈ 0.286 m

Since the maximum x coordinate is reached from the origin, the position vector will be (0.286, 0) meters.

Therefore, the position vector of the particle when it reaches its maximum x coordinate is (0.286, 0) meters.

Do you think it's a good idea to dump your homework questions without showing us you know anything about solving them?

Sure looks like cheating to me.

You've posted seven physics questions in a row.

If you don't understand any of them, then I urge you to drop this class.