A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (-1.00iˆ - 0.500jˆ) m/s2. When it reaches its maximum x coordinate, what are its (a) velocity and (b) position vector?

the position vector (x, y) will be (1.5 m,-2.25 m) and the velocity vector (vx, vy) will be (0 m/s, - 1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = - 1.00 m/s²

x = coordinates in the x-axis

when x reaches its maximum coordinate, then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s) ² + 2 * (-1.00 m/s²) * x

x = 1.5 m

also for the time t

vx = v₀x + ax*t → t = (vx-v₀x) / ax = (0 - 3 m/s) / (-1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = - 0.5 m/s²

y = coordinates in the y-axis

y₀ = initial coordinate in the y-axis = 0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2 * (-0.5 m/s²) * (3 s) ² = - 2.25 m

and

vy=v₀y + ay*t = 0 + (-0.5 m/s²) * (3 s) = (-1.5 m/s)

therefore the position vector (x, y) will be (1.5 m,-2.25 m)

and the velocity vector (vx, vy) will be (0 m/s, - 1.5 m/s)