Chapter 6, Problem 82A

To determine

The speed of the gases relative to the observer.

Answer to Problem 82A

The velocity of the expelled gas with respect to the observer is $-1.5\text{km/s}$ .

Explanation of Solution

Given:

The speed of the rocket relative to the observer is $v_{\text{R/O}}=1.25\text{km/s}$ .

The speed of the expelled gas with respect to the rocket is $v_{\text{G/R}}=1.25\text{km/s}$ .

Consider the speed of the gases with respect to the observer is denoted by $v_{\text{G/O}}$ .

Formula used:

Vector Sum:

Vector additioninvolving relative velocities is

  $v_{\text{a/b}}+v_{\text{b/c}}=v_{\text{a/c}}$

Here, $v_{\text{a/b}}$ is the velocity of $a$ with respect to $b$ , $v_{\text{b/c}}$ is the velocity of $b$ with respect to $c$ , and $v_{\text{a/c}}$ is the velocity of $a$ with respect to $c$ .

Calculation:

Consider the speed of the gases with respect to the observer is denoted by $v_{\text{G/O}}$ .

As the rocket moves forward, the gas is expelled in the backward direction. Hence, the speed of the expelled gas with respect to the rocket is taken as negative, $v_{\text{G/R}}=-2.75\text{km/s}$ .

The speed of the rocket relative to the observer is $v_{\text{R/O}}=1.25\text{km/s}$ .

Apply vector sum,

  $\begin{array}{c}{v}_{\text{G/O}}=v_{\text{G/R}}+v_{\text{R/O}}\\ =-2.75+1.25\\ =-1.5\text{km/s}\end{array}$

Conclusion:

Therefore, the velocity of the expelled gas with respect to the observer is $-1.5\text{km/s}$ .