Question

5 the acceleration vector (15 pts)
(problem 4.2.27)

1. a particles acceleration is (4.0\\(hat{i}\\) + 3.0\\(hat{j}\\)) m/s² at t = 0, its position and velocity are zero.

(a) (10 pts) what are the particles position and velocity as functions of time?
(b) (5 pts) find the equation of the path of the particle, y as a function of x.

Explanation:

Step1: Find velocity as a function of time

The acceleration vector is \(\vec{a}=(4.0\hat{i}+3.0\hat{j})\ m/s^{2}\).
We know that \(\vec{a}=\frac{d\vec{v}}{dt}\).
Integrating the \(x -\) component: \(a_{x} = 4.0\ m/s^{2}\), \(\int_{0}^{v_{x}}dv_{x}=\int_{0}^{t}4.0dt\), \(v_{x}=4.0t\).
Integrating the \(y -\) component: \(a_{y}=3.0\ m/s^{2}\), \(\int_{0}^{v_{y}}dv_{y}=\int_{0}^{t}3.0dt\), \(v_{y}=3.0t\).
So, \(\vec{v}=(4.0t\hat{i}+3.0t\hat{j})\ m/s\).

Step2: Find position as a function of time

We know that \(\vec{v}=\frac{d\vec{r}}{dt}\).
Integrating the \(x -\) component: \(v_{x} = 4.0t\), \(\int_{0}^{x}dx=\int_{0}^{t}4.0t\ dt\), \(x = 2.0t^{2}\).
Integrating the \(y -\) component: \(v_{y}=3.0t\), \(\int_{0}^{y}dy=\int_{0}^{t}3.0t\ dt\), \(y = 1.5t^{2}\).
So, \(\vec{r}=(2.0t^{2}\hat{i}+1.5t^{2}\hat{j})\ m\).

Step3: Find \(y\) as a function of \(x\)

From \(x = 2.0t^{2}\), we get \(t^{2}=\frac{x}{2.0}\).
Substitute \(t^{2}\) into the equation for \(y\): \(y = 1.5t^{2}\), \(y=\frac{1.5}{2.0}x\), \(y = 0.75x\).

Answer:

a. Position: \(\vec{r}=(2.0t^{2}\hat{i}+1.5t^{2}\hat{j})\ m\), Velocity: \(\vec{v}=(4.0t\hat{i}+3.0t\hat{j})\ m/s\)
b. \(y = 0.75x\)