A particle moves in the xy plane with constant acceleration. At time zero, the p

Question

A particle moves in the xy plane with constant acceleration. At time zero, the particle is x=6.5m, y=2.5m, and has velocity Vo = (4m/s)i + (-4m/s)j. The acceleration is given by a= (1.5m/s^2)i + (1.5m/s^2)j.
A. What is the x component of velocity after 4s? Answer in units of m/s
B. What is the y component of velocity after 4s? Answer in units of m/s
C. What is the magnitude of the displacement from the origin (x=0 m, y=0 m) after 4s? Answer in units of m

Explanation / Answer

The particle is moving in xy-plane Poision at time zero = (6.5,2.5) m Velocity at time zero V0= (4i,-4j) m/s => velocity in x direction at time zero = Vi0 = 4m/s => velocity in y direction at time zero = Vj0 = -4m/s Acceleration = a = (1.5i,1.5j) m/s^2 => Acceleration in x direction = ai = 1.5m/s^2 => Acceleration in y direction = aj = 1.5m/s^2 Time = t = 4 s To find the x and v components of velocitis we have totreat i,j components of the velocity and acceleratin separately. Let Vi4,Vj4 be te velocity of the particle in x and y directions respectively Then Vi4 = Vi0 + ai*t Vi4 = 4 + 1.5*4 = 10 m/s The x component of the velocity after 4 seconds is equal to 10 m/s Similarly Vj4 = Vj0 + aj*t Vj4 = -4 + 1.5*4 = 2 m/s The y component of the velocity after 4 seconds is equal to 2 m/s

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