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# A jet airplane travelling at the speed of $500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ ejects its product of combustion at the speed of $1500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ relative to the jet plane. The speed of products of combustion with respect to an observer on the ground isA. $500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$B. $1000\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$C. $1500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$D. $2000\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$

Last updated date: 13th Jun 2024
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Hint:Use the formula for the relative velocity of an object with respect to the other object. Rewrite this formula for the relative velocity of the combustion products of the jet airplane with respect to the speed of the jet airplane. Substitute the values of all the speeds with the correct sign in this formula and calculate the speed of the combustion products with respect to the ground.

Formula used:
The formula for relative speed of an object with respect to the other object is given by
${v_{BA}} = {v_B} - {v_A}$ …… (1)
Here, ${v_{BA}}$ is the relative speed of object B with respect to object A, ${v_B}$ is speed of the object B with respect to ground and ${v_A}$ is the speed of the object A with respect to the ground.

We have given that the speed of the jet airplane with respect to the ground is $500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$.
${v_J} = 500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$
We have also given that the relative speed of the combustion products of the jet airplane with respect to the jet airplane is $1500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$.
${v_{CJ}} = 1500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$
We are asked to calculate the speed of the combustion products of the jet airplane with respect to the ground.

Let ${v_C}$ be the speed of the combustion products of the jet airplane with respect to the ground.We can calculate the speed of the combustion products of the jet airplane with respect to the ground using formula in equation (1). Rewrite equation (1) for the relative speed of the combustion products of the jet airplane with respect to the jet airplane.
${v_{CJ}} = {v_C} - {v_J}$

Rearrange the above equation for ${v_C}$.
${v_C} = {v_{CJ}} + {v_J}$
Substitute $- 1500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ for ${v_{CJ}}$ and $500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ for ${v_J}$ in the above equation.
${v_C} = \left( { - 1500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right) + \left( {500\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right)$
$\therefore {v_C} = - 1000\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$
The negative sign indicates that the combustion products are moving opposite to that of the airplane.Therefore, the speed of the combustion products with respect to the ground is $1000\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$.

Hence, the correct option is B.

Note: The students should not forget to use the value of relative speed of the combustion products of the jet airplane with respect to the jet airplane with negative sign because the combustion products thrown out of the jet airplane are moving in a direction opposite to that of the jet airplane.