Question

10
Driving. In an emergency, the distance,
d m, it takes for a car moving at a
speed of v m/s to stop completely is
given by d = 0.15v2 + v.
(i) When the car has only 80 m to stop,
show that 3v2 + 20v – 1600 < 0.
(ii) Solve the inequality in part (i) to
find the maximum speed of the car.​

Answer 1

Answer:

• i) The proof is below

• ii) The maximum speed is 20m/s

Explanation:

(i) When the car has only 80 m to stop, show that $3v^2+20v-1600<0.$

The distance, $d$ $m$, it takes for a car moving at a speed of $v$ $m/s$  to stop completely is given by $d=0.15v^2+v$

The car must stop before the 80 m, so the distance to stop is less than 80 m:

                       $d<80$

• Substitute:

                      $0.15v^2+v<80$

• Multiply both sides by 20:

             $20\times 0.15v^2+20\times v<20\times 80\\\\3v^2+20v<1600$

• Subtract 1600 from both sides:

            $3v^2+20v-1600<0$

Which is the expression you wanted to prove.

(ii) Solve the inequality in part (i) to find the maximum speed of the car.​

To solve the inequality you need to factor the left side:

• Put 20v as 80v - 60v

         $3v^2+80v-60v-1600<0$

• Use commutative property of addition

         $3v^2-60v+80v-1600<0$

• Use associative property

         $(3v^2-60v)+(80v-1600)<0$

• Factor each binomial:

        $3v(v-20)+80(v-20)<0$

• Common factor $(v-20)$

          $(v-20)(3v+80)<0$

There are two possible solutions: 1) the first factor is positive and the second negative, and 2) the first factor is negative and the second positive.

         

1. First tentative solution

              $v-20>0\\\\v>20\\\\3v+80<0\\\\3v<-80\\\\ v<-80/3\\\\v<-26.26$

The speed cannot be greater than 20m/s and less than - 26.26 m/s at the same tine, thus, there are not solutions for this assumption.

2. Second tentative solution

              $v-20<0\\\\v<20\\\\3v+80>0\\\\3v>-80\\\\ v>-80/3\\\\v>-26.26$

You are looking for an upper bound, thus the solution is that the speed is less than 20m/s.