Answer:
[tex]R_{2}=4R_{1}[/tex]
Explanation:
Let's use the centripetal force definition.
So the first force will be:
[tex]F_{1}=ma_{c}=m\frac{v_{1}^{2}}{R_{1}}[/tex] (1)
Here:
Now, the final speed is double of the first one, so V2 = 2V1, Therefore the force will be:
[tex]F_{2}=m\frac{v_{2}^{2}}{R_{2}}=m\frac{4v_{1}^{2}}{R_{2}}[/tex] (2)
If we match both equation (1) and (2), we will find R2.
[tex]m\frac{v_{1}^{2}}{R_{1}}=m\frac{4v_{1}^{2}}{R_{2}}[/tex]
[tex]\frac{1}{R_{1}}=\frac{4}{R_{2}}[/tex]
Finally,
[tex]R_{2}=4R_{1}[/tex]
I hope it helps you!
The radius of the tightest curve on the same road that it can round without skidding is [tex]4R_{1}[/tex]. Hence, option (B) is correct.
Given data:
The speed of car is, [tex]v_{1}[/tex].
The radius of level curve is, [tex]R_{1}[/tex].
Here the concept of centripetal force is applicable, whose expression is,
[tex]F = \dfrac{mv^{2}_{1}}{R_{1}} .................................................(1)[/tex]
here, m is the mass of car.
Now, the final speed is double of the first one, so [tex]v_{2} = 2v_{1}[/tex]. So, the force will be,
[tex]\leq F' = \dfrac{mv^{2}_{2}}{R_{2}} \\\\F' = \dfrac{m(2v_{1})^{2}}{R_{2}}\\\\F' = \dfrac{4mv_{1}^{2}}{R_{2}} ..........................................................(2)[/tex]
Comparing equation (1) and (2) as,
[tex]F = F'\\\\\dfrac{mv^{2}_{1}}{R_{1}}=\dfrac{4mv_{1}^{2}}{R_{2}}\\\\R_{2} = 4R_{1}[/tex]
Thus, we can conclude that the radius of the tightest curve on the same road that it can round without skidding is [tex]4R_{1}[/tex]. Hence, option (B) is correct.
Learn more about the centripetal force here:
https://brainly.com/question/11324711
