The velocity vefot of a ball is v(t)=-3x+4y at any time. Its initial position is r=12x-4y. The components of the velocity vector are in m/s and the components of the position vector are in meters. The symbols x and y are the unit vetors in x and y directions respectively.
Find the final position vector in component form at the end of 2 seconds.
Find the average acceleration vector.
To find the final position vector, we can integrate the velocity vector with respect to time, starting from the initial position.
The velocity vector v(t) is given as v(t) = -3x + 4y.
We need to integrate the x and y components separately to find the displacement in each direction.
Integrating the x component: ∫(-3x) dx = (-3/2)x^2 + C1
Integrating the y component: ∫(4y) dy = 2y^2 + C2
To find the constants C1 and C2, we need to use the initial position r = 12x - 4y.
Substituting x = 0 and y = 0 into the equations:
(-3/2)(0)^2 + C1 = 12
C1 = 12
2(0)^2 + C2 = -4
C2 = -4
Now we have the integration constants, so we can find the final position vector at the end of 2 seconds.
Plugging in t = 2 into the equations:
Final x component = (-3/2)(2)^2 + 12 = -6 + 12 = 6 meters
Final y component = 2(2)^2 - 4 = 8 - 4 = 4 meters
Therefore, the final position vector in component form at the end of 2 seconds is r = 6x + 4y.
Now let's find the average acceleration vector.
The average acceleration vector can be calculated by dividing the change in velocity by the change in time.
Since we only have the velocity vector equation, we can differentiate it to find the acceleration vector.
Taking the derivative of each component of the velocity vector:
d/dt (-3x) = -3(dx/dt) = -3v_x
d/dt (4y) = 4(dy/dt) = 4v_y
Therefore, the acceleration vector is a = -3v_x + 4v_y.
Note that v_x and v_y are the components of the velocity vector, which in this case are constant (-3 and 4 respectively).
So the average acceleration vector is a = -3(-3) + 4(4) = 9 + 16 = 25 m/s^2.