IDNLearn.com: Your trusted source for finding accurate and reliable answers. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.
Sagot :
Answer:
The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)
Explanation:
Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.
Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].
Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].
Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.
Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.