Aparticle starts from the origin at t=0 with an initial velocity of 5.0 m/s along the positive "x" axis. if the acceleration is: a = (-3.0i+4.5j) m/s ; determine the velocity and position of the particle at the moment it reaches it's maximum "x" coordinate.

The relevant equation to use in this case is:

v = u + at                                                                              ---> 1

where v is final velocity, u is initial velocity, a is acceleration, and t is time

Since u is 5m/s along the positive x axis so u = 5i. We are asked to find the time when maximum x coordinate is reached. This is also the time when the x component of the velocity reaches zero:

v=0

a (at maximum x) = -3.0i

Therefore

0 = 5i - 3i * t

t = 5/3 sec

Maximum x is reached at 5/3 sec

Equation 1 with x&y components is:

v = 5i + (-3i * t + 4.5j * t)                                                ---> 2

Substituting t = 5/3 so get v:

v = 7.5 j m/s

Now we have the velocity, we need to find the position x. We know that:

v = dx / dt

Therefore equation 2 becomes:

dx / dt = 5i +(-3i * t + 4.5j * t)

By integrating we get:

x = (5t)i -((3.5t^2)/2)i +((4.5T^2)/2)j

Now substituting t = 5/3:

x =(125/35)i + (81/100)j

Answers:

v = 7.5 j m/s

x =(125/35)i + (81/100)j

a and b sometimes have a positive value