A certain distance is covered at certain speed if of this distance is covered in double the time the ratio of
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Hi.
Here is the answer---
_______________
Let the Original distance and the time to be covered be x km and y hr respectively.
In First Case,
Using the Formula,
Speed = Distance/Time
Speed (v) = x/y
Now, In second Case,
Time = 2 of Orginal
⇒ Time = 2y
Thus, New Speed(v₁) = x/2y
∴ The Ratio between the Original and the New speed,
= ![\frac{x/y}{x/2y} \frac{x/y}{x/2y}](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%2Fy%7D%7Bx%2F2y%7D+)
![\frac{v}{v1} = \frac{2}{1} \frac{v}{v1} = \frac{2}{1}](https://tex.z-dn.net/?f=+%5Cfrac%7Bv%7D%7Bv1%7D++%3D++%5Cfrac%7B2%7D%7B1%7D+)
Thus, v : v₁ = 2 : 1
Ratio between the Original and the New speed of the Object for the Certain distance is 2 : 1
_________________
Hope it helps.
Have a Nice day.
Here is the answer---
_______________
Let the Original distance and the time to be covered be x km and y hr respectively.
In First Case,
Using the Formula,
Speed = Distance/Time
Speed (v) = x/y
Now, In second Case,
Time = 2 of Orginal
⇒ Time = 2y
Thus, New Speed(v₁) = x/2y
∴ The Ratio between the Original and the New speed,
Thus, v : v₁ = 2 : 1
Ratio between the Original and the New speed of the Object for the Certain distance is 2 : 1
_________________
Hope it helps.
Have a Nice day.
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