A small particle starts from rest from the origin of an xy-coordinate system and travels in the xy-plane. Its acceleration in the x-direction is 2.0 m/s2, and its acceleration in the y-direction is 1.0 m/s2. What is the x-coordinate of the particle when its y-coordinate is 12 m?

The x-coordinate of the particle is 24 m.

Explanation:

In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion

Xf=Xo+Voxt+0.5axt² (I)

Yf=Yo+Voyt+0.5ayt² (II)

Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.

The particle starts from rest from the origin, therefore:

Vox=Voy=0

Xo=Yo=0

Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:

12=0 + (0) t + 0.5 (1.0) t²

12=0.5t²

Dividing by 0.5 and extracting thr squareroot both sides:

t=√12/0.5

t=√24 = 2√6

Replacing t=2√6, ax=2.0, Xo=0 and Vox=0 in (I) to obain the x-coordinate:

Xf=0+0t+0.5 (2.0) (2√6) ²

Xf = 24 m