Question

A rolling ball moves from $x_1=8.4\mathrm{cm}$ to$x_2=-4.2\mathrm{cm}$ during the time from$t_1=3.0\mathrm{s}$ to $t_2=6.1\mathrm{s}$. What is its average velocity over this time interval?

Short answer

The average velocity over the time interval is $-4.1\frac{\mathrm{cm}}{\mathrm{s}}$.

Step-by-step solution

Step 1. Relation between average velocity, positions, and time

The term ‘average velocity’ can be defined as the division of the change in the displacement of a body from one point to another by the time taken in the displacement. It can also be defined as the average displacement divided by the average time taken by an object or a body.

Given data.

The initial position is $x_1=8.4\mathrm{cm}$.

The final position is $x_2=-4.2\mathrm{cm}$.

The initial time is $t_1=3.0\mathrm{s}$.

The final time is $t_2=6.1\mathrm{s}$.

We know that $\mathbf{Average}\mathbf{}\mathbf{velocity}\mathbf{=}\frac{\mathbf{final}\mathbf{}\mathbf{position}\mathbf{-}\mathbf{initial}\mathbf{}\mathbf{position}}{\mathbf{time}\mathbf{}\mathbf{elapsed}}$.

Step 2. Calculation of average velocity using given values

Substituting the values in the above equation,

$\begin{array}{c}\mathrm{Average}\mathrm{velocity}=\frac{\mathrm{final}\mathrm{position}-\mathrm{initial}\mathrm{position}}{\mathrm{time}\mathrm{elapsed}}\\ v=\frac{x_2-x_1}{t_2-t_1}\\ =\frac{-4.2\mathrm{cm}-8.4\mathrm{cm}}{6.1\mathrm{s}-3.0\mathrm{s}}\\ =-4.1\frac{\mathrm{cm}}{\mathrm{s}}\end{array}$

Therefore, the average velocity over the time interval is $-4.1\frac{\mathrm{cm}}{\mathrm{s}}$.