Answer:
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)
