Chapter 2, Problem 1TC

Jogging Johnny runs along a train flatcar that moves at the velocities shown. From greatest to least,rank the relative velocities of Jake as seen by an observer on the ground. (Call the direction to the right positive).

To determine

The ranking of relative velocity from greatest to least of Jake.

Answer to Problem 1TC

The rankingof relative velocity from greatest to least of Jake is $D>C>A>B$

Explanation of Solution

Given info:

The velocity of Jogging Johnny is $4\text{m}/\text{s}$ and the velocity of train flatcar for case A is $8\text{m}/\text{s}$.

Write the expression for the velocity relative to the person on the ground.

$v=v_J+v_T\mathrm{......}(1)$

Here,

$v_J$ is the velocity of Jogging Johnny relative to train flatcar

$v_T$ is the velocity of train flatcar relative to a person on ground

Substitute $-4\text{m}/\text{s}$ for $v_J$ and $+8\text{m}/\text{s}$ for $v_T$ in equation (1) to get $v$ for case A.

$\begin{array}{c}{v}_A=\left(-4\text{m}/\text{s}\right)+\left(+8\text{m}/\text{s}\right)\\ v_A=+4\text{m}/\text{s}\end{array}$

Thus, the relative velocity of Jake as seen by observer on the ground in case A is $v_A=+4\text{m}/\text{s}$ .

Substitute $+6\text{m}/\text{s}$ for $v_J$ and $-10\text{m}/\text{s}$ for $v_T$ in equation (1) to get $v$ for case B.

$\begin{array}{c}{v}_B=\left(+6\text{m}/\text{s}\right)+\left(-10\text{m}/\text{s}\right)\\ v_B=-4\text{m}/\text{s}\end{array}$

Thus, the relative velocity of Jake as seen by observer on the ground in case B is $v_B=-4\text{m}/\text{s}$

Substitute $+4\text{m}/\text{s}$ for $v_J$ and $+6\text{m}/\text{s}$ for $v_T$ in equation (1) to get $v$ for case C.

$\begin{array}{c}{v}_C=\left(+4\text{m}/\text{s}\right)+\left(+6\text{m}/\text{s}\right)\\ v_C=+10\text{m}/\text{s}\end{array}$

Thus, the relative velocity of Jake as seen by observer on the ground in case C is $v_C=+10\text{m}/\text{s}$.

Substitute $-6\text{m}/\text{s}$ for $v_J$ and $+18\text{m}/\text{s}$ for $v_T$ in equation (1) to get $v$ for case D.

$\begin{array}{c}{v}_D=\left(-6\text{m}/\text{s}\right)+\left(+18\text{m}/\text{s}\right)\\ v_D=+12\text{m}/\text{s}\end{array}$

Thus, the relative velocity of Jake as seen by observer on the ground in case D is $v_D=+12\text{m}/\text{s}$

Conclusion:

Therefore, theranking of relative velocity from greatest to least of Jake is $D>C>A>B$ .