Asked by Megan
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.
Answers
Answered by
Damon
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
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
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