Question

The displacement of a particle is given by $\mathrm{x}={(\mathrm{t}-2)}^2$ where $\mathrm{x}$ is in meters and$\mathrm{t}$is in seconds. The distance covered by the particle in first $4$ seconds is?

Answer 1

Step 1: Given data:

The distance traveled is given by $\mathrm{x}={(\mathrm{t}-2)}^2={\mathrm{t}}^2–4\mathrm{t}+4$

Here $\mathrm{x}$ is in meters and $\mathrm{t}$is in seconds.

Step 2: Calculation of velocity:

The velocity can be given as:

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ \Rightarrow \mathrm{v}=\frac{\mathrm{d}({\mathrm{t}}^2–4\mathrm{t}+4)}{\mathrm{dt}} \\ \Rightarrow \mathrm{v}=2\mathrm{t}-4 \end{gathered}$$

Step 3: Calculation of acceleration:

The acceleration can be given as

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \\ \mathrm{a}=\frac{\mathrm{d}\left(2\mathrm{t}-4\right)}{\mathrm{dt}} \\ \mathrm{a}=2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Step 4: Calculation of Distance traveled:

When,

$$\begin{gathered} t=0,v=-4m{s}^{-1} \\ t=2s,v=0m{s}^{-1} \end{gathered}$$
$t=4s,v=4m{s}^{-1}$
Using the given data, the$v-t$ graph can be shown as:

Here, from the given diagram,

The Distance traveled = area of the graph
$$\begin{gathered} =|\mathrm{area}\mathrm{OAC}|+\mathrm{area}\mathrm{ABD} \\ =\mid \frac{4\times 2}{2}\mid +\frac{1}{2}\times 2\times 4 \\ =8\mathrm{units} \end{gathered}$$

Thus, the distance covered by the particle in the first $4$ seconds is $8\mathrm{units}$.