Question

A jet airplane traveling at a speed of $500km/h$ ejects its products of combustion at a speed of $1500km/h$ relative to the jet plane. The speed of the latter with respect to an observer on the ground is:

Answer 1

Step 1: Given data

Speed of jet airplane, $\overrightarrow{v_A}=500km/hr$

Speed of products combusted with respect to the jet plane, $\overrightarrow{v_{BA}}$ $=1500km/hr$

Step 2: Formula used

Relative motion formula $\overrightarrow{v_{BA}}=\overrightarrow{v_B}-\overrightarrow{v_A}$

Step 3: Calculating speed

Relative velocity = Velocity of B -Velocity of A (Vector calculations)

From the diagram we understand, $\overrightarrow{v_{BA}}$ will be negative i.e., $-1500km/hr$.

$\overrightarrow{v_{BA}}=\overrightarrow{v_B}-\overrightarrow{v_A}$

Speed of products with respect to the observer on the ground,

$$\begin{gathered} \overrightarrow{v_B}=-1500+500 \\ =-1000km/hr \end{gathered}$$

The speed of products combusted with respect to the observer standing on the ground is equal to $1000km/h$ in the left direction according to the sign convection.