Q13DQ - University Physics with Modern Physics

Question

The official’s truck in Fig. 2.2 is at at and is at at ${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{277}\mathbf{ }\mathbf{\text{ m}}$, at ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{ s}}$, and is at ${\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{19}\mathbf{ }\mathbf{\text{ m}}$ at ${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{ s}}$. (a) Sketch two different possible x-t graphs for the motion of the truck. (b) Does the average velocity ${\mathit{v}}_{\mathbf{a}\mathbf{v}\mathbf{-}\mathbf{x}}$ during the time interval from ${\mathit{t}}_{\mathbf{1}}$ to ${\mathit{t}}_{\mathbf{2}}$ have the same value for both of your graphs? Why or why not?

Step-by-Step Solution

Answer

(a)

 

The required graphs are:

(b). Yes, the total displacement and time are same for both the graphs.

Given information

It is given that $x_1=277\text{ m}$ at $t_1=16.0\text{ s}$ and$x_2=19\text{ m}$ at $t_2=25\text{ s}$.

Part (a)

Draw the graphs for the given condition:

Part (b)

The formula to find the velocity at different position and at different time is, $v=\frac{x_2-x_1}{t_2-t_1}$.

Use the given values into the above formula:

$\begin{array}{rcl}v& =& \frac{x_2-x_1}{t_2-t_1}\\ & =& \frac{\left(19-277\right)\text{ m}}{\left(25-16\right)\text{ m}}\\ & =& \frac{-258\text{ m}}{9\text{ s}}\\ & =& -28.67\text{ m/s}\end{array}$

As the velocity is negative, then the vehicle is moving in negative direction.

 

Displacement is known as the shortest distance between two points, and in both the graphs in part (a), the shortest distance is same, where the velocity depends on the displacement only, not distance, hence the average velocity is the same for both the paths.