
A train goes at a constant speed. If it covers $150$ miles in $2\dfrac{1}{2}$hours, how long would it take for that train to cover $\left( 1 \right)\,100\,miles\,;$ $\left( 2 \right)\,270\,miles\,;$$\left( 3 \right)\,360\,miles\,$?
Answer
481.5k+ views
Hint: Here the speed is given as constant. Therefore, we can use the formula $Speed = \dfrac{{Dis\tan ce}}{{Time\,taken}}$. Here, distance and time is given in the question, therefore we can find the speed and with the help of this value of speed we can find the time of corresponding distance travelled.
Complete step-by-step answer:
In the above question, it is given as the train is going with a constant speed.
It means the value of acceleration is zero. Therefore, the value of speed does not change throughout its journey.
Therefore, we can use the formula of constant speed here in terms of distance travelled and the time taken as $Speed = \dfrac{{Dis\tan ce}}{{Time\,taken}}$ .
In the above question, it is given as the train covers $150$ miles in $2\dfrac{1}{2}$hours.
We can also write time in an improper fraction as $\dfrac{5}{2}$hours.
Now, using the above formula to find the speed.
$Speed = \dfrac{{Dis\tan ce}}{{Time\,taken}}$
$Speed = \dfrac{{150}}{{\dfrac{5}{2}}}$
$ \Rightarrow Speed = 30 \times 2$
On simplification, we get
$Speed = 60\,miles/hr$
Now, we will find the time taken for a given distance travelled.
$\left( 1 \right)\,100\,miles\,$
We can also write the above formula as,
$Time\,Taken = \dfrac{{Dis\tan ce}}{{Speed}}$
On putting the values of distance and speed in the above formula, we get
$ \Rightarrow Time\,Taken = \dfrac{{100}}{{60}}$
$Time\,Taken = \dfrac{5}{3}\,hr$
$\left( 2 \right)\,270\,miles\,$
We can also write the above formula as,
$Time\,Taken = \dfrac{{Dis\tan ce}}{{Speed}}$
On putting the values of distance and speed in the above formula, we get
$ \Rightarrow Time\,Taken = \dfrac{{270}}{{60}}$
$Time\,Taken = \dfrac{9}{2}\,hr$
$\left( 3 \right)\,360\,miles\,$
We can also write the above formula as,
$Time\,Taken = \dfrac{{Dis\tan ce}}{{Speed}}$
On putting the values of distance and speed in the above formula, we get
$ \Rightarrow Time\,Taken = \dfrac{{360}}{{60}}$
$Time\,Taken = 6\,hr$
Note: The speed is the scalar quantity. The speed does not specify the direction of the object. If the objects are moving at a constant speed it refers to the objects covering an equal length of distance at the same time. Acceleration is defined as the ratio of velocity and time. The constant body speed cannot be accelerated. When the speed of an object is the same then it's called the constant speed. Here the speed does not increase or decrease.
Complete step-by-step answer:
In the above question, it is given as the train is going with a constant speed.
It means the value of acceleration is zero. Therefore, the value of speed does not change throughout its journey.
Therefore, we can use the formula of constant speed here in terms of distance travelled and the time taken as $Speed = \dfrac{{Dis\tan ce}}{{Time\,taken}}$ .
In the above question, it is given as the train covers $150$ miles in $2\dfrac{1}{2}$hours.
We can also write time in an improper fraction as $\dfrac{5}{2}$hours.
Now, using the above formula to find the speed.
$Speed = \dfrac{{Dis\tan ce}}{{Time\,taken}}$
$Speed = \dfrac{{150}}{{\dfrac{5}{2}}}$
$ \Rightarrow Speed = 30 \times 2$
On simplification, we get
$Speed = 60\,miles/hr$
Now, we will find the time taken for a given distance travelled.
$\left( 1 \right)\,100\,miles\,$
We can also write the above formula as,
$Time\,Taken = \dfrac{{Dis\tan ce}}{{Speed}}$
On putting the values of distance and speed in the above formula, we get
$ \Rightarrow Time\,Taken = \dfrac{{100}}{{60}}$
$Time\,Taken = \dfrac{5}{3}\,hr$
$\left( 2 \right)\,270\,miles\,$
We can also write the above formula as,
$Time\,Taken = \dfrac{{Dis\tan ce}}{{Speed}}$
On putting the values of distance and speed in the above formula, we get
$ \Rightarrow Time\,Taken = \dfrac{{270}}{{60}}$
$Time\,Taken = \dfrac{9}{2}\,hr$
$\left( 3 \right)\,360\,miles\,$
We can also write the above formula as,
$Time\,Taken = \dfrac{{Dis\tan ce}}{{Speed}}$
On putting the values of distance and speed in the above formula, we get
$ \Rightarrow Time\,Taken = \dfrac{{360}}{{60}}$
$Time\,Taken = 6\,hr$
Note: The speed is the scalar quantity. The speed does not specify the direction of the object. If the objects are moving at a constant speed it refers to the objects covering an equal length of distance at the same time. Acceleration is defined as the ratio of velocity and time. The constant body speed cannot be accelerated. When the speed of an object is the same then it's called the constant speed. Here the speed does not increase or decrease.
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