
A $ 100 $ meter long train is running at the speed of $ 30km/hr $ . How much time is taken by it to pass a man standing near the railway line?
(A) $ 10\sec $
(B) $ 11\sec $
(C) $ 12\sec $
(D) $ 20\sec $
Answer
539.7k+ views
Hint :For solving this particular question, we have to use the relationship between metres and kilometres that is $ 1km = 1000m $ , we just have to apply $ v = \dfrac{d}{t} $
Where ,
‘v’ is the speed in meters per second ,
‘d’ is the distance travelled by train in meters ,
And ‘t’ is the time taken by train to cover the distance in seconds .
Complete Step By Step Answer:
It is given that, the speed of train , $ v = 30km/hr $ ,
And the length of the train, $ d = 100m...............(1) $
For crossing a man who is standing near the railway line, the train has to cover a distance equal its length, that is , $ d = 100m $ .
As we know that speed is given as distance travelled per unit time .
Mathematically, we can write ,
$ v = \dfrac{d}{t} $
Where ,
‘v’ is the speed in meters per second ,
‘d’ is the distance travelled by train in meters ,
And ‘t’ is the time taken by train to cover the distance in seconds .
We get ,
$ t = \dfrac{d}{v} $
Both the distance and speed should be in their corresponding standard unit.
We have distance already in its standard unit.
Now, we are converting speed into its standard unit ,
$ v = 30km/hr = \dfrac{{30 \times 1000}}{{60 \times 60}}m/\sec $
$ = \dfrac{{50}}{6}m/\sec ................(2) $
Now, substitute $ (1),(2) $ in $ t = \dfrac{d}{v} $ ,
We will get the following,
$ t = \dfrac{d}{v} = \dfrac{{100}}{{\dfrac{{50}}{6}}} $ sec
Simply the above expression, we will get ,
$ t = \dfrac{{100 \times 6}}{{50}} $ sec
Simply the above expression, we will get ,
\[\therefore t = 12\sec \]
Therefore, we can say that option ‘C’ is the correct option.
Note :
In the given question, no mathematical formula except the relation $ 1km = 1000m $ and $ v = \dfrac{d}{t} $
Where ,
‘v’ is the speed in meters per second ,
‘d’ is the distance travelled by train in meters ,
And ‘t’ is the time taken by train to cover the distance in seconds .
is being used only the mathematical operations such as addition, subtraction, multiplication and division is used. Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.
Where ,
‘v’ is the speed in meters per second ,
‘d’ is the distance travelled by train in meters ,
And ‘t’ is the time taken by train to cover the distance in seconds .
Complete Step By Step Answer:
It is given that, the speed of train , $ v = 30km/hr $ ,
And the length of the train, $ d = 100m...............(1) $
For crossing a man who is standing near the railway line, the train has to cover a distance equal its length, that is , $ d = 100m $ .
As we know that speed is given as distance travelled per unit time .
Mathematically, we can write ,
$ v = \dfrac{d}{t} $
Where ,
‘v’ is the speed in meters per second ,
‘d’ is the distance travelled by train in meters ,
And ‘t’ is the time taken by train to cover the distance in seconds .
We get ,
$ t = \dfrac{d}{v} $
Both the distance and speed should be in their corresponding standard unit.
We have distance already in its standard unit.
Now, we are converting speed into its standard unit ,
$ v = 30km/hr = \dfrac{{30 \times 1000}}{{60 \times 60}}m/\sec $
$ = \dfrac{{50}}{6}m/\sec ................(2) $
Now, substitute $ (1),(2) $ in $ t = \dfrac{d}{v} $ ,
We will get the following,
$ t = \dfrac{d}{v} = \dfrac{{100}}{{\dfrac{{50}}{6}}} $ sec
Simply the above expression, we will get ,
$ t = \dfrac{{100 \times 6}}{{50}} $ sec
Simply the above expression, we will get ,
\[\therefore t = 12\sec \]
Therefore, we can say that option ‘C’ is the correct option.
Note :
In the given question, no mathematical formula except the relation $ 1km = 1000m $ and $ v = \dfrac{d}{t} $
Where ,
‘v’ is the speed in meters per second ,
‘d’ is the distance travelled by train in meters ,
And ‘t’ is the time taken by train to cover the distance in seconds .
is being used only the mathematical operations such as addition, subtraction, multiplication and division is used. Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.
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