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A bird is tossing (flying to and fro) between two cards moving towards each other on a straight road. One car has a speed of $18 \mathrm{~km} / \mathrm{h}$ while the other has the speed of $27 \mathrm{~km} /$ $h$. The bird starts moving from first car towards the other and is moving with the speed of $36 \mathrm{~km} / \mathrm{h}$ and when the two cars were separated by $36 \mathrm{~km}$. What is the total distance covered by the bird?
Solution:
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Verified Answer
As given that, speed of first car $=18 \mathrm{~km} / \mathrm{h}$
Speed of second car $=27 \mathrm{~km} / \mathrm{h}$
So, relative speed of each car $=18+27=45 \mathrm{~km} / \mathrm{h}$ and distance between the cars $=36 \mathrm{~km}$
Then, time of meeting the cars $(t)$
$$
\begin{aligned}
&=\frac{\text { Distance between the cars }}{\text { Relative speed of cars }} \\
&=\frac{36}{45}=\frac{4}{5}=0.8 \text { hours }
\end{aligned}
$$
Given, speed of the bird $\left(v_0\right)=36 \mathrm{~km} / \mathrm{h}$
$\therefore$ Distance covered by the bird in $(0.8)$ hours
$=v_b \times t=36 \times 0.8=28.8 \mathrm{~km}$
Speed of second car $=27 \mathrm{~km} / \mathrm{h}$
So, relative speed of each car $=18+27=45 \mathrm{~km} / \mathrm{h}$ and distance between the cars $=36 \mathrm{~km}$
Then, time of meeting the cars $(t)$
$$
\begin{aligned}
&=\frac{\text { Distance between the cars }}{\text { Relative speed of cars }} \\
&=\frac{36}{45}=\frac{4}{5}=0.8 \text { hours }
\end{aligned}
$$
Given, speed of the bird $\left(v_0\right)=36 \mathrm{~km} / \mathrm{h}$
$\therefore$ Distance covered by the bird in $(0.8)$ hours
$=v_b \times t=36 \times 0.8=28.8 \mathrm{~km}$
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