Step-by-step explanation:
The approximate stopping distance d(in feet) is given by the formula :
[tex]d=0.05v^2+2.2v[/tex] ...(1)
Where
v is the speed of the car in mph
We need to find the speed of the car when the distance is 200 feet
Put d = 200 in equation (1)
[tex]0.05v^2+2.2v=200\\\\0.05v^2+2.2v-200=0[/tex]
It is a quadratic equation. Divide the above equation by 0.05.
[tex]v^2+44v-4000=0[/tex]
The solution of the above equation is given by :
[tex]v=\dfrac{-44\pm \sqrt{44^2- 4(1)(-4000)} }{2(1)}\\\\v=\dfrac{-44+ \sqrt{44^2- 4(1)(-4000)} }{2(1)},\dfrac{-44- \sqrt{44^2- 4(1)(-4000)} }{2(1)}\\\\v=44.96\ ft/h, -88.96\ ft/h[/tex]
Since, 1 mph = 5280 ft/hour
44.96 ft/h = 0.00851 mph
-88.96 = -0.0168 mph
Hence, this is the required siolution.
