Question

The position of a particle in the xy-plane at time t is r(t)=(t+4)i+(t ^2 −5) j. Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at t=3. The equation for the path of the particle is y=

Answer 1

The equation for the path of the particle is
`\(y = t^2 - 5\)`. At t = 3, the particle's velocity vector is
`\(\mathbf{v}(3) = \mathbf{i} + 6\mathbf{j}\)`, and the acceleration vector is
`\(\mathbf{a}(3) = \mathbf{0} + 2\mathbf{j}\)`.

Step-by-step explanation:

To find the equation for the path of the particle, we use the position vector
`\( \mathbf{r}(t) = (t + 4)\mathbf{i} + (t^2 - 5)\mathbf{j} \).` The y-coordinate of this vector is
`\(y = t^2 - 5\)`, giving us the equation for the path of the particle.

To find the velocity vector, we take the derivative of the position vector with respect to time t. The velocity vector
`\( \mathbf{v}(t) \)` is given by
`\( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t\mathbf{j} \)`. At t = 3, the velocity vector is
`\( \mathbf{v}(3) = \mathbf{i} + 6\mathbf{j} \).`

Similarly, to find the acceleration vector, we take the derivative of the velocity vector with respect to time. The acceleration vector
`\( \mathbf{a}`(t) is given by
`\( \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \mathbf{0} + 2\mathbf{j} \`). At \(t = 3\), the acceleration vector is
`\( \mathbf{a}(3) = 2\mathbf{j} \).`

In summary, the equation for the path of the particle is
`\(y = t^2 - 5\)`, and at
`\(t = 3\)`, the particle's velocity vector is
`\( \mathbf{v}(3) = \mathbf{i} + 6\mathbf{j} \)`, and the acceleration vector is
`\( \mathbf{a}(3) = 2\mathbf{j} \).`