A particle in motion

Initial approach

We have a particle that moves in a single dimension, in the x axis, as it is deduced from the position equation. Therefore, we can dispense with the vector notation and calculate all quantities as scalars. If we are asked for the corresponding vector quantities, we simply multiply the scalar magnitude by the unit vector $\overrightarrow{i}$ . In any case, we illustrate the process in each section. It should be remembered that in one dimension the position vector will be given by $\overrightarrow{r}\left(t\right)=x\left(t\right)\overrightarrow{i}+\cancel{y\left(t\right)\overrightarrow{j}}+\cancel{z\left(t\right)\overrightarrow{k}}$.

1. The displacement in the x axis is given by  Δx:

$$∆x=x_f-x_i=x\left(4\right)-x\left(3\right)=\left(4^2-5\left(4\right)+1.2\right)-\left(3^2-5\left(3\right)+1.2\right)=2 \text{m}$$

Rigorously, the displacement is a vector magnitude and will be given by:

$$∆\overrightarrow{r}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i=∆x\overrightarrow{i}+\cancel{∆y\overrightarrow{j}}+\cancel{∆z\overrightarrow{k}}=\boxed{2\overrightarrow{i} \text{m}}$$

For the calculation of the average velocity in the x axis, divide the displacement by the time employed:

$$v_{av{g}_x}=\frac{∆x}{∆ t}=\frac{2}{1}=2 \text{m/s}$$

And rigorously, using the vector notation the average velocity will be given by:

$${\overrightarrow{v}}_{avg}=v_{av{g}_x}\overrightarrow{i}+\cancel{v_{av{g}_y}\overrightarrow{j}}+\cancel{v_{av{g}_z}\overrightarrow{k}}=\boxed{2\overrightarrow{i} \text{m/s}}$$

2. Following a similar procedure, we can obtain a general formula for the calculation of the displacement on the x axis. For that, we substitute t + Δt (final time) and t (initial time):

$$∆x=x_f-x_i=x\left(t+∆t\right)-x\left(t\right)=\left((t+∆t{)}^2-5(t+∆t)+1.2\right)-\left(t^2-5\left(t\right)+1.2\right)=∆t^2-5∆t+2t∆t =∆t(∆t+2t-5)\text{m}$$

However, once more, if we want to be rigorous we must apply vector notation, using the general formula of the displacement vector of the particle

$$∆\overrightarrow{r}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i=∆x\overrightarrow{i}+\cancel{∆y\overrightarrow{j}}+\cancel{∆z\overrightarrow{k}}=\boxed{∆t(∆t+2t-5)\text{m}}$$

We have considered t as initial time and t + Δt as the final time. If we consider the values of the previous section, t = 3 s and Δt = 1 s and replace them in the calculated general equation of the displacement vector, we obtain the same value.

3. 3. The calculation of the velocity at any given time can be made by taking the derivative of the position vector. In this case:

$$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$$

In this case, we are only interested in the x axis, and we get:

$$v_x=\frac{dx\left(t\right)}{dt}=2t-5 \text{m/s}$$

However, we are asked to do the calculation using limit:

$$v_x=\underset{∆t\to 0}{\lim }\frac{∆x}{∆ t}=\frac{\cancel{∆t}(∆t+2t-5)}{\cancel{∆t}}=2t-5 \text{m/s}$$

In any case, being rigorous and using the vector notation, the velocity vector is:

$$\overrightarrow{v}\left(t\right)=v_x\left(t\right)\overrightarrow{i}+\cancel{v_y\left(t\right)\overrightarrow{j}}+\cancel{v_z\left(t\right)\overrightarrow{k}}=\boxed{(2t-5)\overrightarrow{i}\text{m/s}}$$