
Applying the brakes can stop a train in certain distance. When such a braking force is made one fourth, the brakes will stop the exact train in a distance, which is given as,
A. same
B. four times
C. double
D. half
Answer
561k+ views
Hint: Braking force is the force which is provided by the brakes in order to stop the train. Greater the braking force, shorter will be the distance covered by the train after the application of brakes. This means that the braking force is inversely proportional to the stopping distance. Apply this condition in the question and compare the values. Hope these ll may help you to solve this question.
Complete step by step answer:
As we all know, the braking force applied will be inversely proportional to the distance covered by the train for stopping. This can be written mathematically as,
$F\propto \dfrac{1}{s}$
Where $F$ be the braking force applied and $s$be the distance covered by the train after the application of brakes.
Now let us look at the parameters mentioned in the question. Let us assume that the braking force in the first condition will be taken as
${{F}_{1}}=F$
And the distance of stopping after the application of this force will be
$s={{s}_{1}}$
The braking force in the second condition is mentioned as the quarter of the force in the first case. That is,
${{F}_{2}}=\dfrac{F}{4}$
And the stopping distance can be taken as,
$s={{s}_{2}}$
Substituting all these values in the equation of the relation between the force and stopping distance will give,
$\begin{align}
& F\propto \dfrac{1}{s} \\
& \dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{{{s}_{2}}}{{{s}_{1}}} \\
\end{align}$
Substituting the values,
\[\dfrac{F}{\dfrac{F}{4}}=\dfrac{{{s}_{2}}}{{{s}_{1}}}\]
That is from this we can say that,
\[\dfrac{{{s}_{2}}}{{{s}_{1}}}=4\]
This means that the stopping distance of the train in the second condition will be four times greater than that of the first condition.
\[{{s}_{2}}=4\times {{s}_{1}}\]
So, the correct answer is “Option B”.
Note: When a force is applied to the brakes of a car or a train, there will be work done by the friction between the brakes and the wheel. If the vehicle travels faster, the braking force needed to stop it at a certain distance will be greater. A greater braking force will create a greater deceleration.
Complete step by step answer:
As we all know, the braking force applied will be inversely proportional to the distance covered by the train for stopping. This can be written mathematically as,
$F\propto \dfrac{1}{s}$
Where $F$ be the braking force applied and $s$be the distance covered by the train after the application of brakes.
Now let us look at the parameters mentioned in the question. Let us assume that the braking force in the first condition will be taken as
${{F}_{1}}=F$
And the distance of stopping after the application of this force will be
$s={{s}_{1}}$
The braking force in the second condition is mentioned as the quarter of the force in the first case. That is,
${{F}_{2}}=\dfrac{F}{4}$
And the stopping distance can be taken as,
$s={{s}_{2}}$
Substituting all these values in the equation of the relation between the force and stopping distance will give,
$\begin{align}
& F\propto \dfrac{1}{s} \\
& \dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{{{s}_{2}}}{{{s}_{1}}} \\
\end{align}$
Substituting the values,
\[\dfrac{F}{\dfrac{F}{4}}=\dfrac{{{s}_{2}}}{{{s}_{1}}}\]
That is from this we can say that,
\[\dfrac{{{s}_{2}}}{{{s}_{1}}}=4\]
This means that the stopping distance of the train in the second condition will be four times greater than that of the first condition.
\[{{s}_{2}}=4\times {{s}_{1}}\]
So, the correct answer is “Option B”.
Note: When a force is applied to the brakes of a car or a train, there will be work done by the friction between the brakes and the wheel. If the vehicle travels faster, the braking force needed to stop it at a certain distance will be greater. A greater braking force will create a greater deceleration.
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