Question

PLEASE HELP!!!!

Answer 1

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let $u$ denote the initial velocity of the vehicle ($20\; \text{mph}$ or $60\; \text{mph}$) and let $v$ denote the velocity of the vehicle after braking ($0\; \text{mph}$). Let $x$ denote the braking distance.

Assume that the acceleration during braking are both constantly $a$ in both scenarios. The SUVAT equations would apply. In particular:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}$.

Since $v = 0$ (the vehicle has completely stopped), the equation becomes $x = (-u^{2}) / (2\, a)$.

Assuming that $a$ (braking acceleration) stays the same, the braking distance $x$ would be proportional to $u^{2}$, the square of the initial velocity.

Hence, increasing the initial speed from $20\; \text{mph}$ to $60\; \text{mph}$ would increase the braking distance by a factor of $3^{2} = 9$.

Answer 2

Answer:

9 times

Explanation: