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

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

i think its c if i'm wrong