Motion in Two Dimensions-Motion in a Plane - Equations of Motion in a Plane with Constant Acceleration

When an object moves under constant acceleration in two dimensions, its motion can be treated as two separate one-dimensional motions along the x- and y-axes.

1. Acceleration Definition
   \[\vec{a}=\frac{\vec{v}-\vec{u}}{t}\]
2. Velocity
   \[\vec v\] = \[\vec u\] + \[\vec a\]t
   Components: \[v_x=u_x+a_xt,\quad v_y=u_y+a_yt\]
3. Displacement
   Average velocity: \[\vec{v}_{\mathrm{avg}}=\frac{\vec{u}+\vec{v}}{2}\]
   Thus, 
   \[\vec{s}=\vec{v}_\mathrm{avg~}t=\vec{u}t+\frac{1}{2}\vec{a}t^2\]
   Components: \[s_x=u_xt+\frac{1}{2}a_xt^2,\quad s_y=u_yt+\frac{1}{2}a_yt^2\]

Given:

\[\vec{u}=5\hat{i}+10\hat{j}\mathrm{m/s},\quad\vec{a}=2\hat{i}+3\hat{j}\mathrm{m/s}^2,t=5\mathrm{s}.\]

Velocity
\[\vec{v}=\left(5+2\times5\right)\hat{i}+\left(10+3\times5\right)\hat{j}=15\hat{i}+25\hat{j}\mathrm{m/s}\]

Magnitude:
v = \[\sqrt{15^2+25^2}=29.15\mathrm{m/s},\theta=\tan^{-1}\left(\frac{25}{15}\right)=59^\circ.\]

Displacement
\[\vec{s}=(5\cdot5+\frac{1}{2}\cdot2\cdot5^{2})\hat{i}+(10\cdot5+\frac{1}{2}\cdot3\cdot5^{2})\hat{j}=50\hat{i}+87.5\hat{j}\mathrm{m}.\]

Magnitude:
s = \[\sqrt{50^{2}+87.5^{2}}=100.78\mathrm{m},\phi=\tan^{-1}\left(\frac{87.5}{50}\right)=60^{\circ}15^{\prime}.\]

Imagine kicking a soccer ball:

• Horizontal motion (x-axis): constant speed (no horizontal acceleration).
• Vertical motion (y-axis): slows downward due to gravity (vertical acceleration).