A particle leaves the origin with an initial velocity v = (3.00 i hat) m/s and a constant acceleration a = (-3.00 i hat - 1.400 j hat) m/s2

(a) What is the velocity of the particle when it reaches its maximum x coordinate?

m/s i hat + m/s j hat + m/s k hat

(b) What is the position vector of the particle at this time?

m i hat + m j hat + m k hat

a)  —0.5 j m/s  

b) 4.5 i + 2.25 j m

Explanation:

Givens:

v_0 =3.00 i m/s

a= (-3 i — 1.400 j ) m/s^2  

The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :

v_x = v_0 + at = 0  

(3.00 i m/s) + (-3 i m/s^2)t=0

t = (3 m/s)/-3 i m/s^2

t = -1 s

Therefore the particle reaches the maximum x-coordinate at time t = 1 s.  

Part a The velocity-of course- is all in the y-direction,therefore:

v_y =v_0+ at

We have that v_0 = 0 in the y-direction.

v_y = (-0.5 j m/s^2)(1 s)

      = —0.5 j m/s  

Part b: While the position of the particle at t = 1 s is given by:

r=r_0+v_0*t+1/2*a*t^2

Where r_0 = 0 since the particle started from the origin.

Its position at t = 1 s is then given by :

r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2

 =4.5 i + 2.25 j m

i would say the answer is food.

hope this !