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Question

# Find the position vector-valued function $\mathbf { r } ( t )$, given that $\mathbf { a } ( t ) = \mathbf { i } + e ^ { t } \mathbf { j }$, $\mathbf { v } ( 0 ) = 2 \mathbf { j }$, and $\mathbf { r } ( 0 ) = 2 \mathbf { i }$.

Solution

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The acceleration of the object is given by $\mathbf{a}(t)=\mathbf{i}+e^t\mathbf{j}$

Let, $\mathbf{r}(t)$ is the object's position at any time.

Therefore,

\begin{aligned} &\mathbf{a}(t)=\mathbf{i}+e^t\mathbf{j}\\ \implies &\dfrac{d}{dt}(\mathbf{v}(t))=\mathbf{i}+e^t\mathbf{j} \end{aligned}

Integrating both sides of the above equation we get,

\begin{aligned} &\int\dfrac{d}{dt}(\mathbf{v}(t))=\int[\mathbf{i}+e^t\mathbf{j} ]dt\\\\ \implies&\dfrac{d}{dt}(\mathbf{r}(t))=\int dt~\mathbf{i}+\int e^tdt~ \mathbf{j}\\\\ \implies&\mathbf{v}(t)=t\mathbf{i}+e^t\mathbf{j}+c_1, \end{aligned}

where $c_1$ is the integrating constant vector.

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