Answer to Question #147396 in Algebra for Rafay

Question

The braking distance d of a car traveling v mi/hr is approximated by d=v+(v^2/20). Determine velocities that result in braking distance of less than 5ft.

Answer 1

To begin with, we need to recognize given valuables:

d=5 ft = 0,00094697 mi and also the equation which connects velocity and distance.

d=v+(v²/20).

So if we put the given length or given breaking distance into equation, we will find the velocity that the car can stop in less than 5 ft.

0,00094697 mi = v+($\frac{v^2}{20}$),

v²+20$\times$ v - 20$\times$ 0.00094697=0

we will solve this equation by finding its roots:

but first we need to compute discriminant then we can find roots:

D=b²-4$\times a \times c$ ,

D=20²+4$\times 1\times0.0189394$ =400.07557

there are two roots and one of them is negative and another of them is positive. We will find positive one:

v=$\frac{-20+\sqrt{\smash[b]{D}}}{2}$ =0.000946 $\frac{mi}{hr}$

if we compute previous equation, we will reach:

v=0.000946 $\frac{mi}{hr}$ .

if the car move with v then it can stop less than 5 ft.