A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration?

Question

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a = (5.00 m/s^2)i + (7.00 m/s^2)j. At time t = 0, the velocity is (4.00 m/s)i. What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 12.0 m. parallel to the x axis?

Answer 1

Attack plan:
To solve this problem we will make use of two of the equations of motion. Then we will find 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 (12m) i:

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


===Part 1 - solving for the time it takes to travel 12m i
We do so by finding the final velocity using (1), and then plugging that back into (2).

v_fi^2 = 4^2 + 2*5*12
v_fi^2 = 16 + 120 = 136

v_fi = 11.662 m/s ((()))

plugging this into (2)

11.662 = 4 + 5*t_f
5*t _f= 7.662 (i just switched sides to make things pretty after i subtracted)

t_f = 7.662 / 5

t_f = 1.532 seconds

===Part 2 - solving for the vertical velocity

v_fj = v_0j + at
v_fj = 0 + 7*1.532

v_fj = 10.724 m/s ((()))

===Part 3 - finding magnitude and angle

magnitude is just

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

V = 15.843 m/s

Angle
Theta = arctan(v_fj/v_fi)
Theta = 0.744 radians

converting radians to degrees (180/pi)

Theta = .744(180/pi) deg= 42.628 deg

Answer:
V = 15.843 m/s
Theta = 42.628 deg