Question

an accepted relationship between stopping distance, d in feet, and the speed of a car, in mph, is d(v)=1.2v+o.o4v^2 on dry, level concrete.
a) how many feet will it take a car traveling 50 mph to stop on dry, level concrete?
b) if an accident occurs 250 feet ahead, what is the maximum speed at which one can travel to avoid being involved in the accident?

Answer 1

so what happens if you put 50 in for v in the equation

Answer 2

it would be 160

Answer 3

how would you do b??

Answer 4

\[d(v) = 250 = 1.2v+0.04v^2\]Can you solve that for\(v\)?

Answer 5

are we solving for v?

Answer 6

You can use the quadratic formula to find the solutions of a polynomial\(ax^2+bx+c=0, a\ne0\):
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
One of the solutions won't make sense here.

Answer 7

Yes, solve for \(v\).

Answer 8

would the answer be -.04 +- square root of 1200.0016)/2.4

Answer 9

What are your values of a, b, and c?

Answer 10

opps i mean -1.2 +- squareroot of (41.44)/.08

Answer 11

What part of that are you dividing by 0.08?

Answer 12

all of them

Answer 13

That's what I figured, but that's not what you wrote. What you wrote is
\[-1.2\pm \frac{\sqrt{41.44}}{0.08}\]

Answer 14

What are your numeric answer?

Answer 15

answers

Answer 16

\[(-1.2 + \sqrt{41.44})/0.08 =\]
\[(-1.2-\sqrt{41.44})/0.08 = \]

Answer 17

The negative value of \(v\) doesn't make sense as an answer in this context, so we'll discard it.

Answer 18

first would be -95.47
second would be 65.47
would the answer be the second one since it is the max.

Answer 19

Yes, the second one is the only one that makes sense, and that's the correct answer.