A particle's constant acceleration is south at 2.0 m/s2. At t = 0, its velocity is 40.0 m/s east. What is its velocity at t = 7.7 s?
Magnitude m/s
Direction
Please someone help me out on this...Thank you
Acceleration is normal to the velocity v(E); therefore, this is centripetal (normal) acceleration. It doesn’t change “east-directed” velocity v(E). The velocity in south direction is
v(S) =a•t = 2•7.7 = 15.4 m/s.
v =sqrt[v(S)²+v(E)²] =
= sqrt[15.4² +40²] =42.86 m/s.
Direction:
tan α =v(S)/v(E) =15.4/40 =0.385.
α = 21º (with East-direction)