
How long will a boy sitting near the window of a train travelling at \[54\;{\rm{km/h}}\]see a train passing by in the opposite direction with a speed of \[36\;{\rm{km/h}}\]? The length of the slow-moving train is \[100\;{\rm{m}}\].
Answer
509.4k+ views
Hint:The above problem can be resolved by applying the formula for the relative speed between each train. Then the relative speed's obtained value is to calculate the value of time, with the given distance as the length of one of the trains. This value of time is the required time by which one train passes the other train. Moreover, the concept of relative speed along with the sign convention is of greater significance. As, the bodies moving along the same direction are taken as positive, while the bodies in the opposite direction are taken as negative.
Complete step by step answer:
Given:
The speed of fast-moving trains is, \[{v_1} = 54\;{\rm{km/h}} = 54\;{\rm{km/h}} \times \dfrac{{\dfrac{5}{{18}}\;{\rm{m/s}}}}{{1\;{\rm{km/h}}}} = 15\;{\rm{m/s}}\].
The speed of slow-moving trains is, \[{v_2} = 54\;{\rm{km/h}} = 36\;{\rm{km/h}} \times \dfrac{{\dfrac{5}{{18}}\;{\rm{m/s}}}}{{1\;{\rm{km/h}}}} = 10\;{\rm{m/s}}\].
The length of slow- moving train is, \[d = 100\;{\rm{m}}\].
The relative speed of both the train is given as,
\[\begin{array}{c}
{v_3} = {v_1} - \left( { - {v_2}} \right)\\
= {v_1} + {v_2}
\end{array}\]
Substituting the values as,
\[\begin{array}{l}
{v_3} = 15\;{\rm{km/h}} + 10\;{\rm{km/h}}\\
{{\rm{v}}_3} = 25\;{\rm{km/h}}
\end{array}\]
The time required to pass the train is,
\[\begin{array}{l}
t = \dfrac{d}{{{v_3}}}\\
t = \dfrac{{100\;{\rm{m}}}}{{25\;{\rm{m/s}}}}\\
t = 4\;{\rm{s}}
\end{array}\].
Therefore, the time required to pass the train is 4 seconds.
Note: Try to understand the relative speed concept and apply the formula of relative speed to resolve the given condition. Moreover, the significant relation taking the relative speed, the required time is calculated. Besides, the average speed and distance concept can also be utilized to obtain the desired result. In addition, the concept of the relative speed can be more significantly understood by taking the direction as the major aspect for the calculation.
Complete step by step answer:
Given:
The speed of fast-moving trains is, \[{v_1} = 54\;{\rm{km/h}} = 54\;{\rm{km/h}} \times \dfrac{{\dfrac{5}{{18}}\;{\rm{m/s}}}}{{1\;{\rm{km/h}}}} = 15\;{\rm{m/s}}\].
The speed of slow-moving trains is, \[{v_2} = 54\;{\rm{km/h}} = 36\;{\rm{km/h}} \times \dfrac{{\dfrac{5}{{18}}\;{\rm{m/s}}}}{{1\;{\rm{km/h}}}} = 10\;{\rm{m/s}}\].
The length of slow- moving train is, \[d = 100\;{\rm{m}}\].
The relative speed of both the train is given as,
\[\begin{array}{c}
{v_3} = {v_1} - \left( { - {v_2}} \right)\\
= {v_1} + {v_2}
\end{array}\]
Substituting the values as,
\[\begin{array}{l}
{v_3} = 15\;{\rm{km/h}} + 10\;{\rm{km/h}}\\
{{\rm{v}}_3} = 25\;{\rm{km/h}}
\end{array}\]
The time required to pass the train is,
\[\begin{array}{l}
t = \dfrac{d}{{{v_3}}}\\
t = \dfrac{{100\;{\rm{m}}}}{{25\;{\rm{m/s}}}}\\
t = 4\;{\rm{s}}
\end{array}\].
Therefore, the time required to pass the train is 4 seconds.
Note: Try to understand the relative speed concept and apply the formula of relative speed to resolve the given condition. Moreover, the significant relation taking the relative speed, the required time is calculated. Besides, the average speed and distance concept can also be utilized to obtain the desired result. In addition, the concept of the relative speed can be more significantly understood by taking the direction as the major aspect for the calculation.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Net gain of ATP in glycolysis a 6 b 2 c 4 d 8 class 11 biology CBSE

Give two reasons to justify a Water at room temperature class 11 chemistry CBSE
