Answer
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Hint: Here we will find the speed of the first train by using the relative speed formula. First, we will assume the speed of the first train to be \[x\] and find the relative speed of the first train with respect to the second train. Then we will convert our all value in one unit. Finally we will use the formula of speed to get the required answer.
Complete step-by-step answer:
Let us take the speed of the first train as \[x\] \[{\rm{km/hr}}\].
The speed of second train \[ = 30{\rm{km/hr}}\]
So, the relative speed of first train with respect to second \[ = \left( {x - 30} \right){\rm{km/hr}}\]
Converting the above value in m/sec we will multiply it with \[\dfrac{5}{{18}}\]. Therefore, we get
\[ \Rightarrow \] The relative speed of first train with respect to second \[ = \left( {x - 30} \right) \times \dfrac{5}{{18}}{\rm{m/sec}}\]
Length of first train is 66 m and second train is 88 m.
So, Total distance travelled \[ = \left( {66 + 88} \right){\rm{m}} = 154{\rm{m}}\]
Time taken \[ = 0.168\min = \left( {0.168 \times 60} \right)\sec \]
Multiplying the terms, we get
\[ \Rightarrow \] Time taken \[ = 10.08\sec \]
Substituting the above values in the formula Speed \[ = \] Distance \[ \div \] Time, we get
\[\left( {x - 30} \right) \times \dfrac{5}{{18}} = \dfrac{{154}}{{10.08}}\]
Multiplying \[\dfrac{{18}}{5}\] on both the sides, we get
\[\begin{array}{l} \Rightarrow \left( {x - 30} \right) = \dfrac{{154}}{{10.08}} \times \dfrac{{18}}{5}\\ \Rightarrow \left( {x - 30} \right) = 55\end{array}\]
Adding 30 on both the sides, we get
\[ \Rightarrow x = 55 + 30\]
\[ \Rightarrow x = 85\]
So, the speed of the first train is \[85{\rm{km/hr}}\].
Hence, option (A) is correct.
Note:
“Relative” is also known as “in comparison to”. The relative speed concept is used when two or more bodies are moving with some speed considered. The relative speed of two bodies is added if they are moving in the opposite direction and subtracted if they are moving in the same direction. The speed of the moving body when considered with respect to the speed of the stationary body is known as relative speed.
Complete step-by-step answer:
Let us take the speed of the first train as \[x\] \[{\rm{km/hr}}\].
The speed of second train \[ = 30{\rm{km/hr}}\]
So, the relative speed of first train with respect to second \[ = \left( {x - 30} \right){\rm{km/hr}}\]
Converting the above value in m/sec we will multiply it with \[\dfrac{5}{{18}}\]. Therefore, we get
\[ \Rightarrow \] The relative speed of first train with respect to second \[ = \left( {x - 30} \right) \times \dfrac{5}{{18}}{\rm{m/sec}}\]
Length of first train is 66 m and second train is 88 m.
So, Total distance travelled \[ = \left( {66 + 88} \right){\rm{m}} = 154{\rm{m}}\]
Time taken \[ = 0.168\min = \left( {0.168 \times 60} \right)\sec \]
Multiplying the terms, we get
\[ \Rightarrow \] Time taken \[ = 10.08\sec \]
Substituting the above values in the formula Speed \[ = \] Distance \[ \div \] Time, we get
\[\left( {x - 30} \right) \times \dfrac{5}{{18}} = \dfrac{{154}}{{10.08}}\]
Multiplying \[\dfrac{{18}}{5}\] on both the sides, we get
\[\begin{array}{l} \Rightarrow \left( {x - 30} \right) = \dfrac{{154}}{{10.08}} \times \dfrac{{18}}{5}\\ \Rightarrow \left( {x - 30} \right) = 55\end{array}\]
Adding 30 on both the sides, we get
\[ \Rightarrow x = 55 + 30\]
\[ \Rightarrow x = 85\]
So, the speed of the first train is \[85{\rm{km/hr}}\].
Hence, option (A) is correct.
Note:
“Relative” is also known as “in comparison to”. The relative speed concept is used when two or more bodies are moving with some speed considered. The relative speed of two bodies is added if they are moving in the opposite direction and subtracted if they are moving in the same direction. The speed of the moving body when considered with respect to the speed of the stationary body is known as relative speed.
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