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