Solved problem: Velocity starting with the position equation

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Statement

difficulty

If a body moves according to the following equation:

$\vec{r (t)} = (4 \cdot t + t^{2}) \cdot \vec{i} + 4 \cdot t \cdot \vec{j} m$

Calculate its instantaneous velocity at time t=1 s.

Solution

Data

$\vec{r (t)} = (4 \cdot t + t^{2}) \cdot \vec{i} + 4 \cdot t \cdot \vec{j} m$

Resolution

To solve this problem, we will use the following equation, that establishes that the instantaneous velocity is the derivative of the position vector with respect to time.

\vec{v} = \lim_{∆ t \to 0} {\vec{v}}_{a v g} = \lim_{∆ t \to 0} \frac{∆ \vec{r}}{∆ t} = \frac{d \vec{r}}{d t}

Differentiating or calculating the limit, we get that:

$\vec{v (t)} = (4 + 2 \cdot t) \cdot \vec{i} + 4 \cdot \vec{j} m$

Once we know the instantaneous velocity vector, we substitute the value of t=1 s and we get the instantaneous velocity for said time:

$\vec{v (1)} = (4 + 2 \cdot 1) \cdot \vec{i} + 4 \cdot \vec{j} m \Rightarrow \vec{v (1)} = 6 \cdot \vec{i} + 4 \cdot \vec{j} m$

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas

Related sections

\vec{v} = \lim_{∆ t \to 0} {\vec{v}}_{a v g} = \lim_{∆ t \to 0} \frac{∆ \vec{r}}{∆ t} = \frac{d \vec{r}}{d t}

Instantaneous Velocity

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Velocity starting with the position equation

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