What is the x and y component of velocity after 4s?
A particle moves in the x-y plane with constant acceleration. At time zero, the particle is at x = 6.5 m, y = 2.5 m, 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.
The car initially is at x = 6.5 m and y = 2.5 m and its velocity is Vo= (4m/s)i +(-4m/s)j.
The acceleration acting on the particle is a=(1.5m/s^2)i + (1.5m/s^2)j. The component of this acceleration acting on the x-component of the velocity is 1.5 m/s^2 and the component of the velocity acting on the y-component of the velocity is also 1.5 m/s^2.
In 4s, the x-component of the velocity due to the constant acceleration changes from 4 m/s to 4 + 1.5*4 = 10 m/s. And the y-component of the velocity due to the constant acceleration changes from -4 m/s to -4 + 1.5*4 = 2 m/s
After 4 s the car has a velocity with an x-component of 10 m/s and a y=component of 2 m/s.
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
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
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
The position at time zero has no effect on the velocity of the particle but will affect its final position.