Answer:
Explanation:
(i) When the car has only 80 m to stop, show that [tex]3v^2+20v-1600<0.[/tex]
The distance, [tex]d[/tex] [tex]m[/tex], it takes for a car moving at a speed of [tex]v[/tex] [tex]m/s[/tex] to stop completely is given by [tex]d=0.15v^2+v[/tex]
The car must stop before the 80 m, so the distance to stop is less than 80 m:
[tex]d<80[/tex]
[tex]0.15v^2+v<80[/tex]
[tex]20\times 0.15v^2+20\times v<20\times 80\\\\3v^2+20v<1600[/tex]
[tex]3v^2+20v-1600<0[/tex]
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:
[tex]3v^2+80v-60v-1600<0[/tex]
[tex]3v^2-60v+80v-1600<0[/tex]
[tex](3v^2-60v)+(80v-1600)<0[/tex]
[tex]3v(v-20)+80(v-20)<0[/tex]
[tex](v-20)(3v+80)<0[/tex]
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
[tex]v-20>0\\\\v>20\\\\3v+80<0\\\\3v<-80\\\\ v<-80/3\\\\v<-26.26[/tex]
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
[tex]v-20<0\\\\v<20\\\\3v+80>0\\\\3v>-80\\\\ v>-80/3\\\\v>-26.26[/tex]
You are looking for an upper bound, thus the solution is that the speed is less than 20m/s.