
A train is moving with a speed of $40km\,h{{r}^{-1}}$. As soon as another train moving in the opposite direction passes by the window, the passenger of the first train starts his stopwatch and notes that the other train passes the window is $3s$. Find the speed of the train moving in the opposite direction is its length is $75m$.
Answer
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Hint: Two trains are moving opposite each other. The length of the second train will be covered relatively quickly due to relative velocity. The speed of a body is related to distance covered and time taken, using this relation and the given values we can calculate the relative velocity and then calculate the real velocity of second train by substituting calculated relative velocity and velocity of first train in the formula for relative velocity.
Formula used:
$s=\dfrac{d}{t}$
${{v}_{r}}={{v}_{1}}-{{v}_{2}}$
Complete solution:
Given, the speed of first train is
$\begin{align}
& 40km\,h{{r}^{-1}}=40\times \dfrac{1000}{3600} \\
& \Rightarrow 40km\,h{{r}^{-1}}=11.1m{{s}^{-1}} \\
\end{align}$
Therefore, the speed of the first train is $11.1m{{s}^{-1}}$, time taken to completely pass the second train is $3s$, the length of the second train is $75m$. Let the relative velocity of the second train be $x$.
We know that,
$s=\dfrac{d}{t}$
Here, $s$ is the speed
$d$ is the distance travelled
$t$ is the time taken
In the above equation, we substitute given values to get,
$\begin{align}
& x=\dfrac{75}{3} \\
& \Rightarrow x=25m{{s}^{-1}} \\
\end{align}$
Therefore, the relative velocity is $25m{{s}^{-1}}$.
The relative velocity is calculated as
${{v}_{r}}={{v}_{1}}-{{v}_{2}}$
Here, ${{v}_{r}}$ is the relative velocity
${{v}_{1}}$ is velocity of the first train
${{v}_{2}}$ is the velocity of the second train
In the above equation, we substitute given values to get,
$\begin{align}
& 25=11.1-(-x) \\
& \Rightarrow 25=11.1+x \\
& \Rightarrow 25-11.1=x \\
& \therefore x=13.9m{{s}^{-1}} \\
\end{align}$
Therefore, the velocity of the second train is $13.9m{{s}^{-1}}$.
Note:
The velocity of the second train is moving opposite to the first, that is why its velocity is taken negative. The relative velocity is the velocity of one body with respect to the other. As both trains are passing by one another, the length of the train will be covered in less time and the relative velocity will be greater.
Formula used:
$s=\dfrac{d}{t}$
${{v}_{r}}={{v}_{1}}-{{v}_{2}}$
Complete solution:
Given, the speed of first train is
$\begin{align}
& 40km\,h{{r}^{-1}}=40\times \dfrac{1000}{3600} \\
& \Rightarrow 40km\,h{{r}^{-1}}=11.1m{{s}^{-1}} \\
\end{align}$
Therefore, the speed of the first train is $11.1m{{s}^{-1}}$, time taken to completely pass the second train is $3s$, the length of the second train is $75m$. Let the relative velocity of the second train be $x$.
We know that,
$s=\dfrac{d}{t}$
Here, $s$ is the speed
$d$ is the distance travelled
$t$ is the time taken
In the above equation, we substitute given values to get,
$\begin{align}
& x=\dfrac{75}{3} \\
& \Rightarrow x=25m{{s}^{-1}} \\
\end{align}$
Therefore, the relative velocity is $25m{{s}^{-1}}$.
The relative velocity is calculated as
${{v}_{r}}={{v}_{1}}-{{v}_{2}}$
Here, ${{v}_{r}}$ is the relative velocity
${{v}_{1}}$ is velocity of the first train
${{v}_{2}}$ is the velocity of the second train
In the above equation, we substitute given values to get,
$\begin{align}
& 25=11.1-(-x) \\
& \Rightarrow 25=11.1+x \\
& \Rightarrow 25-11.1=x \\
& \therefore x=13.9m{{s}^{-1}} \\
\end{align}$
Therefore, the velocity of the second train is $13.9m{{s}^{-1}}$.
Note:
The velocity of the second train is moving opposite to the first, that is why its velocity is taken negative. The relative velocity is the velocity of one body with respect to the other. As both trains are passing by one another, the length of the train will be covered in less time and the relative velocity will be greater.
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