A moving train 66 m long overtakes another train 88 m long moving in the same direction in \[0.168\min \]. If the second train is moving at 30 km/hr, at what speed is the first train moving?
A. 85 \[{\rm{km/hr}}\]
B. 50 \[{\rm{km/hr}}\]
C. 55 \[{\rm{km/hr}}\]
D. 25 \[{\rm{km/hr}}\]
Answer
573.3k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, what is the term for a bowler taking five wickets in an innings?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

What is deficiency disease class 10 biology CBSE

