Chapter 2.4, Problem 38SSC

To determine

To Rank: The position-time graphs according to the average speed from greatest average speed to least average speed.

Answer to Problem 38SSC

  $v_A>v_B>v_C>v_D$

Explanation of Solution

Given info:

The position-time graphs with different speeds.

Formula used:

The average velocity of a particle is given by

  $\overline{v}=\frac{\Delta d}{\Delta t}=\frac{d_{final}-d_{initial}}{t_{final}-t_{initial}}$

Calculation:

Assume that each box in the position-time graph indicates 1 unit.

Now, it is known that the magnitude of average velocity is the average speed only. Thus, following this concept and above formula of average velocity, different speeds of plots A, B, C, and D will be,

  $\begin{array}{c}{v}_A=\left|{\overline{v}}_{\text{A}}\right|=\left|\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}\right|\\ =\left|\frac{\left(-3-3\right)\text{m}}{\left(3-0\right)\text{s}}\right|\\ =2\text{m/s}\end{array}$

  $\begin{array}{c}{v}_B=\left|{\overline{v}}_{\text{B}}\right|=\left|\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}\right|\\ =\left|\frac{\left(5-0\right)\text{m}}{\left(3.4-0\right)\text{s}}\right|\left(\text{approx}\right)\\ =1.47\text{m/s}\end{array}$

  $\begin{array}{c}{v}_C=\left|{\overline{v}}_{\text{C}}\right|=\left|\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}\right|\\ =\left|\frac{\left(-3-2\right)\text{m}}{\left(5-0\right)\text{s}}\right|\\ =1\text{m/s}\end{array}$

And,

  $\begin{array}{c}{v}_D=\left|{\overline{v}}_{\text{D}}\right|=\left|\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}\right|\\ =\left|\frac{\left(4-1\right)\text{m}}{\left(5-0\right)\text{s}}\right|\\ =0.6\text{m/s}\end{array}$

Conclusion:

Thus, using the above calculated values,

  $v_A>v_B>v_C>v_D$