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

Solution

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Answered 8 months ago
Answered 8 months ago
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The acceleration of the object is given by a(t)=i+etj\mathbf{a}(t)=\mathbf{i}+e^t\mathbf{j}

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

Therefore,

a(t)=i+etj    ddt(v(t))=i+etj\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,

ddt(v(t))=[i+etj]dt    ddt(r(t))=dt i+etdt j    v(t)=ti+etj+c1,\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 c1c_1 is the integrating constant vector.

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