Answer
Velocity: \[ \mathbf{v}(t) = (2 + t)\mathbf{i} + (2 - e^{-t})\mathbf{j} \] Position: \[ \mathbf{r}(t) = \left(\frac{t^2}{2} + 2t + 1\right)\mathbf{i} + \left(2t + e^{-t} - 2\right)\mathbf{j} \]
Work Step by Step
Step 1 Given: \[ \mathbf{a}(t) = \mathbf{i} + e^{-t}\mathbf{j} \] with initial conditions: \[ \mathbf{v}(0) = 2\mathbf{i} + \mathbf{j}, \quad \mathbf{r}(0) = \mathbf{i} - \mathbf{j} \] Step 2 To find the velocity: \[ \mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int (\mathbf{i} + e^{-t}\mathbf{j}) \, dt \] \[ = t\mathbf{i} - e^{-t}\mathbf{j} + \mathbf{c} \] Since \(\mathbf{v}(0) = 2\mathbf{i} + \mathbf{j}\), then: \[ \mathbf{c} = 2\mathbf{i} + 2\mathbf{j} \] So, \[ \mathbf{v}(t) = (2 + t)\mathbf{i} + (2 - e^{-t})\mathbf{j} \] Step 3 To find the position: \[ \mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \int ((2 + t)\mathbf{i} + (2 - e^{-t})\mathbf{j}) \, dt \] \[ = \left(\frac{t^2}{2} + 2t + 1\right)\mathbf{i} + \left(2t + e^{-t} - 2\right)\mathbf{j} + \mathbf{c}_1 \] Since \(\mathbf{r}(0) = \mathbf{i} - \mathbf{j}\), then: \[ \mathbf{c}_1 = \mathbf{i} - \mathbf{j} \] So, \[ \mathbf{r}(t) = \left(\frac{t^2}{2} + 2t + 1\right)\mathbf{i} + \left(2t + e^{-t} - 2\right)\mathbf{j} \] Result Velocity: \[ \mathbf{v}(t) = (2 + t)\mathbf{i} + (2 - e^{-t})\mathbf{j} \] Position: \[ \mathbf{r}(t) = \left(\frac{t^2}{2} + 2t + 1\right)\mathbf{i} + \left(2t + e^{-t} - 2\right)\mathbf{j} \]
