Particle Motion with Constant Acceleration

Particle Motion with Constant Acceleration

Example 4.5: Motion in x-y plane with constant acceleration

x (m)
y (m)
0.0 s

Results at t = 0.0 s:

Position: (0.0 m, 0.0 m)

Velocity: (5.0 m/s, 0.0 m/s)

Speed: 5.0 m/s

Physics Explanation:

This simulation demonstrates the motion of a particle with initial velocity and constant acceleration in the x-y plane.

Given:

  • Initial velocity: \( \mathbf{v}_0 = 5.0 \hat{\mathbf{i}} \, \text{m/s} \)
  • Constant acceleration: \( \mathbf{a} = (3.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}}) \, \text{m/s}^2 \)

Position as function of time:

\[ \mathbf{r}(t) = \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 \]

\[ x(t) = 5.0 t + 1.5 t^2 \]

\[ y(t) = 1.0 t^2 \]

Velocity as function of time:

\[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = (5.0 + 3.0 t) \hat{\mathbf{i}} + 2.0 t \hat{\mathbf{j}} \]

Textbook Questions:

(a) When x = 84 m:

\[ 5.0 t + 1.5 t^2 = 84 \Rightarrow t = 6 \, \text{s} \]

\[ y(6) = 1.0 \times 6^2 = 36.0 \, \text{m} \]

(b) Speed at t = 6 s:

\[ \mathbf{v}(6) = 23.0 \hat{\mathbf{i}} + 12.0 \hat{\mathbf{j}} \, \text{m/s} \]

\[ \text{Speed} = \sqrt{23^2 + 12^2} \approx 26 \, \text{m/s} \]

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