A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration ModifyingAbove a With right-arrow equals left-parenthesis 5.0ModifyingAbove i With caret plus 8.0ModifyingAbove j With caret right-parenthesis m/s squared. At time t = 0, the velocity is left-parenthesis 5.0ModifyingAbove i With caret right-parenthesis m/s. What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

To answer this problem we will make use of two of the equations of motion. Then we will come across out each velocity component. The difficult part comes in solving for the j component of velocity for which we will have to know t_f, the time it takes to be displaced (11m):

Our equations
1. v_f^2 = v_0^2 + 2ax

2. v_f = v_0 + at

Solving for the time it takes to travel 11m

We do so by finding the final velocity using (1), and then plugging that back into (2).

v_fi^2 = 4^2 + 2*5*11

v_fi^2 = 16 + 110 = 126

v_fi = 11.22 m/s ((()))

plugging this into (2)

11.22 = 4 + 5*t_f

5*t _f= 7.22

t_f = 7.22 / 5

t_f = 1.444 seconds

Solving for the vertical velocity

v_fj = v_0j + at

v_fj = 0 + 7*1.444

v_fj = 10.108 m/s ((()))

Finding magnitude and angle

V = sqrt(v_fi^2 + v_fj^2)

125.884 + 10.108

V = 15.10151 m/s

Angle:

Theta = arctan(v_fj/v_fi)

Theta = 0.0801 radians

converting radians to degrees (180/pi)

Theta = .0.0801(180/pi) deg= 4.591 deg

Answer:

V = 15.10151 m/s

Theta = 4.591 deg