Answer:
Magnitude the of the acceleration is 3.32[tex]m/s^2[/tex] and direction is south
Explanation:
[tex]v_{0} =70.9 m/s\\v=14 m/s\\S=727 m[/tex]
we know that
[tex]v^2 = v_{0}^2 +2aS[/tex]
by substituting the values we can get the required acceleration
[tex]v^2 = v_{0}^2 +2aS\\14^2 = 70.9^2 +2\times a\times 727\\a=3.32 m/s^2[/tex]
Magnitude the of the acceleration is 3.32[tex]m/s^2[/tex] and direction is south
