Question

A jet airplane at the speed of $500\mathrm{km}·{\mathrm{h}}^{-1}$ ejects its products of combustion at the speed of $1500\mathrm{km}·{\mathrm{h}}^{-1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

• A. $500\mathrm{km}·{\mathrm{h}}^{-1}$
• B. $-1000\mathrm{km}·{\mathrm{h}}^{-1}$
• C. $1500\mathrm{km}·{\mathrm{h}}^{-1}$
• D. $2000\mathrm{km}·{\mathrm{h}}^{-1}$

Answer 1

The correct option is B

$-1000\mathrm{km}·{\mathrm{h}}^{-1}$

Step 1: Given

The jet is moving with a velocity: $\overrightarrow{V_j}=500\mathrm{km}·{\mathrm{h}}^{-1}$

Velocity of the product of combustion with respect to the plane is: $\overrightarrow{{\mathrm{V}}_{\mathrm{gj}}}=-1500km·h^{-1}$
For an observer, the velocity is: $V_o=0km·h^{-1}$

Let us assume that the velocity of the product of combustion with respect to the observer is $\overrightarrow{V_g}$

Step 2: Formula Used

The relative velocity of an object $A$ with respect to object $B$ is given by the formula $V_{AB}=V_A-V_B$, where $V_A$ is the velocity of $A$ and $V_B$ is the velocity of $B$.

Step 3: Solution

The difference of the velocity of jet from the velocity of gas will give the relative velocity of the gas with respect to jet.

$$\begin{gathered} \overrightarrow{V_{gj}}=\overrightarrow{V_g}-\overrightarrow{V_j} \\ \Rightarrow -1500=\overrightarrow{V_g}-500 \\ \Rightarrow \overrightarrow{V_g}=-1500+500 \\ \Rightarrow \overrightarrow{V_g}=-1000 \end{gathered}$$

Therefore, the velocity of the latter with respect to an observer on the ground is $-1000\mathrm{km}·{\mathrm{h}}^{-1}$.

Hence, the correct option is (B).