An ice skater is gliding along at 2.1 m/s , when she undergoes an acceleration of magnitude 1.4 m/s2 for 3.0 s. After that she's moving at 6.3 m/s . Find the angle between her acceleration vector and her initial velocity vector.

Answer 1

oh well

original velocity = 2.1 i + 0 j

final velocity = Vx i + Vy j

sqrt (Vx^2 + Vy^2) = 6.3
so
Vx^2 + Vy^2 = 39.7

(Vx - 2.1)/3 = Ax
(Vy - 0)/3 = Ay

so Ay = Vy/3
Ax^2 + Ay^2 = 1.4^2 = 1.96
so three equations
Ax^2 + Vy^2/9 = 1.96
Vx^2+Vy^2 = 39.7
Ax = (Vx - 2.1)/3