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A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of $27{\text{km}}{{\text{h}}^{ - 1}}$ while the other car has a speed of $18{\text{km}}{{\text{h}}^{ - 1}}$. The bird starts moving from the first car towards the other and is moving with the speed of $36{\text{km}}{{\text{h}}^{ - 1}}$ when the two were separated by $36{\text{km}}$. Find the total distance covered by the bird.
A) $28 \cdot 8{\text{km}}$
B) $38 \cdot 8{\text{km}}$
C) $48 \cdot 8{\text{km}}$
D) $58 \cdot 8{\text{km}}$

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Last updated date: 20th Jun 2024
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Answer
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Hint:Here it is not only the bird that moves with some speed, but it is also the two cars that are moving with some speed. So the relative velocity of the cars is what matters. The bird will continue to toss between the two cars until they meet. So if we were to obtain the time taken for the two cars to meet, then the total distance covered by the bird can be calculated.

Formulas used:
-The relative velocity of two bodies is given by, ${v_{rel}} = {v_1} - {v_2}$ where ${v_1}$ is the velocity of the first body and ${v_2}$ is the velocity of the second body.
-The time taken by a body to cover a distance is given by, $t = \dfrac{s}{v}$ where $s$ is the distance covered and $v$ is the speed of the body.
-The distance covered by a body is given by, $s = v \times t$ where $v$ is the speed of the body and $t$ is the time taken to cover the distance.

Complete step by step solution.
Step 1: List the parameters known from the question.
The problem at hand involves two cars moving towards each other and a bird tossing between the first car and the second car.
Now, the speed of the first car is given to be ${v_1} = 27{\text{km}}{{\text{h}}^{ - 1}}$ .
The speed of the second car is given to be ${v_2} = 18{\text{km}}{{\text{h}}^{ - 1}}$ .
The speed of the bird is given to be ${v_b} = 36{\text{km}}{{\text{h}}^{ - 1}}$ .
The distance between the two cars when the bird starts to move is given to be ${s_1} = 36{\text{km}}$ .
Let the total distance covered by the bird in time $t$ be ${s_{total}}$ .

Step 2: Express the relation for the relative speed of the two cars to find the time taken for the two cars to meet.
The relative velocity of the two cars is given by, ${v_{rel}} = {v_1} - {v_2}$ ----------- (1)
Substituting for ${v_1} = 27{\text{km}}{{\text{h}}^{ - 1}}$ and ${v_2} = - 18{\text{km}}{{\text{h}}^{ - 1}}$ in equation (1) we get, ${v_{rel}} = 27 - \left( { - 18} \right) = 45{\text{km}}{{\text{h}}^{ - 1}}$ .
So the relative velocity of the two cars is ${v_{rel}} = 45{\text{km}}{{\text{h}}^{ - 1}}$ .
The time taken for the two cars to meet when they are ${s_1}$ distance apart can be expressed as $t = \dfrac{{{s_1}}}{{{v_{rel}}}}$ -------- (2)
Substituting for ${s_1} = 36{\text{km}}$ and ${v_{rel}} = 45{\text{km}}{{\text{h}}^{ - 1}}$ in equation (2) we get $t = \dfrac{{36}}{{45}} = \dfrac{4}{5}{\text{h}}$
So the time taken for the two cars to meet is $t = \dfrac{4}{5}{\text{h}}$. This is how long the bird continues to toss between the two cars.

Step 3: Express the relation for the total distance covered by the bird in the obtained time $t$ .
The total distance covered by the bird in time $t$ can be expressed as ${s_{total}} = {v_b} \times t$ -------- (3)
Substituting for $t = \dfrac{4}{5}{\text{h}}$ and ${v_b} = 36{\text{km}}{{\text{h}}^{ - 1}}$ in equation (3) we get, ${s_{total}} = 36 \times \dfrac{4}{5} = 28 \cdot 8{\text{km}}$
So the total distance covered by the bird is ${s_{total}} = 28 \cdot 8{\text{km}}$ .

Hence the correct option is A.

Note:The two cars are mentioned to be moving towards each other. This suggests that they are moving in opposite directions. If we take the first car to be moving in the positive x-direction, then the second car would be moving in the negative x-direction. So we substitute ${v_2} = - 18{\text{km}}{{\text{h}}^{ - 1}}$ in equation (1).