A particle with with velocity Vo = -2i + 4j m/s at t=0 undergoes a constant acceleration ã of magnitude 3m/s² at an angle ø equals 130 from the positive direction at x axis, what is the particle of the velocity V at t=s in unit vector notation as a magnitude and angle
Answers
Given : A particle with with velocity Vo = -2i + 4j m/s at t=0 undergoes a constant acceleration ã of magnitude 3m/s² at an angle ø equals 130 from the positive direction at x axis,
To Find : what is the particle of the velocity V at t=5s in unit vector notation as a magnitude and angle
Solution:
Vo = -2i + 4j at t=0
acceleration ã of magnitude 3m/s² at an angle ø equals 130 from the positive direction at x axis
=> a = 3Cos130°i + 3Sin130° j
=> a = -3Cos50° i + 3Sin50°j
=> a = -1.928i + 2.298j
Vt = Vo + at
t = 5
=> V = -2i + 4j + (-1.928i + 2.298j) * 5
=> V = -2i + 4j - 9.64i + 11.49j
=> V = -11.64i + 15.49j
Magnitude = √ (-11.64)² + (15.49)² = 19.738
Tanα = 15.49/(-11.64)
=> Tanα = -1.33
=> α = 127°
(19.738 , ∠127°)
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