Question

IP A particle passes through the origin with a velocity of (6.8m/s)y^.

Part A If the particle's acceleration is (−5.0m/s2)x^, what are its x and y positions after 5.0 s ?

x, y =

  m

Part B

If the particle's acceleration is (−5.0m/s2)x^, what are vx and vy after 5.0 s ?

Express your answers using two significant figures separated by a comma.

vx, vy =

  m/s  

Answer 1

S = So + Vo*t + a*t²/2
being so = (0,0)m (origin), vo = (0, 6.8) m/s

(because the velocity of the body is 6.8 m/s in y direction and in x direction is 0 m/s)

and a = (-5.0, 0) m/s². so:

s = (0,0) + (0, 6.8)t + (-5.0/2 , 0)t²
s = (-2.5t² , 6.8t)
so, applying t = 5.0s, we have s = (-62.5 , 17) m.

b) applying in the equation for the velocity as funtion of time, we have:
v = vo + at
v = (0, 6.8) + (-5.0 , 0)t
v = (-5t, 6.8)
applying t = 5.0 S,

we have:
v = (-25.0 , 6.8) m/s