Question

A jet airplane traveling at a speed of $500{\rm{ }}{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ ejects its products of combustion at the speed of $150{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ relative to the jet plane. What is the speed of combustion products w.r.t. an observer on the ground?

Answer 1

Hint:This question is based on Relative Velocity principle. If the velocity of an object A is defined from the point of reference of another object B, then the relative velocity of object A w.r.t. object B is given by,
\[{\vec V_{AB}} = {\vec V_A} + {\vec V_B}\]
Where, \[{\vec V_A}\] represents the velocity vector of object A
\[{\vec V_B}\] represents the velocity vector of object B, and
\[{\vec V_{AB}}\] represents the relative velocity vector of object A w.r.t. object B.

Complete step by step answer:
Given:
The velocity of jet airplane \[{V_1} = 500\;{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}\]
The velocity of the products of combustion ${V_2} = - 150\;{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}$
Negative sign (-) indicates that the direction of the products of combustion is opposite to the direction of the jet airplane.
Now using the relative velocity formula, the velocity of the combustion products w.r.t. an observer on the ground denoted by ${V_r}$ is given by-
${V_r} = {V_1} + {V_2}$
Substituting the values of \[{V_1}\] and ${V_2}$ in the expression, we get,
\[\begin{array}{l}
{V_r} = 500 + \left( { - 150} \right)\\
\Rightarrow {V_r} = 500 - 150\\
\Rightarrow {V_r} = 350\;{{{\rm{km}}} {\left/
 {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
 } {{\rm{hr}}}}
\end{array}\]
Therefore, the speed of combustion products w.r.t. an observer on the ground is \[350\;{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.
} {{\rm{hr}}}}\].

Note: It should be noted that in order to find the speed of combustion products relative to an observer on the ground, the speed of the observer is zero, which means that the point of reference is fixed.
It means that,
The speed of combustion products w.r.t. an observer on the ground $\begin{array}{l}= {V_r} + {\rm{ the speed of the observer}}\\
{\rm{ = }}{V_r} + 0\\= {V_r}\end{array}$
And we know that \[{V_r} = 350\;{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}\]