position/velocity?

Question

A particle leaves the origin with an initial velocity v = 3.00 i, in meters per second. It experiences a constant acceleration a = -2.00 i - 1.400 j, in meters per second squared.

(a) What is the velocity of the particle when it reaches its maximum x coordinate?
______m/s i + _______m/s j + __________m/s k

(b) What is the position vector of the particle at this time?
_______m i + _________m j + __________m k

Answer 1

vi = voi + ai * t
0 = 3m/s + -2m/s^2 * t
t=1.5s
x = xo + 1/2 * (vio + vi) * t
x = 0 + 1/2 * (3m/s + 0) * 1.5s
x = 2.25m
y = yo + voj * t + 1/2 * aj * t^2
y = 0 + 0 + 1/2 * -1.4m/s * (1.5s)^2
y = -1.575m
vj = voj + aj * t
vj = 0 + 1.4m/s^2 * 1.5s
vj = -2.1m/s

(a) 0m/s i + -2.1m/s j + 0m/s k
(b) 2.25m i + -1.575m j + 0m k

It has been a few years since I have used these equations please check my work

Answer 2

I forget, is i along x or y? I'll guess that it's along x.

Thanks to Trevor O, who showed me something I missed. And I saw something Trevor O missed. So I think, between us, this is right now.

a. The x axis acceleration of -2i is slowing the particle in x motion. When the x velocity decreases to zero, it starts going in the -x direction.
Vf = Vo + a*t
When it reaches the max x, Vf=0. Look like the time to come to a stop would be 1.5 s.

So in that time, the velocity in y would be
-1.4*t = 2.1 m/s
__0__m/s i + __2.1__m/s j + __0__m/s k

b. In x you can calculate the average velocity.
(3i + 0)/2 = 1.5 m/s in the x direction

So the x component of the position vector would be
(1.5 m/s)*1.5 s = 2.25i in m/s

In y
y = (1/2)(-1.4 m/s^2*t^2
y = -1.575j

So position vector is
__2.25__m i + __-1.575__m j + __0__m k