Question

When a driver needs to stop a car, the approximate stopping distance d(in feet) is given by the formula: d=0.05v^(2)+2.2v, where v is the speed of the car (in miles per hour). Suppose a car travels 200 feet before stopping(d=200). How fast was the car traveling?

Answer 1

Step-by-step explanation:

The approximate stopping distance d(in feet) is given by the formula :

$d=0.05v^2+2.2v$ ...(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)

$0.05v^2+2.2v=200\\\\0.05v^2+2.2v-200=0$

It is a quadratic equation. Divide the above equation by 0.05.

$v^2+44v-4000=0$

The solution of the above equation is given by :

$v=\dfrac{-44\pm \sqrt{44^2- 4(1)(-4000)} }{2(1)}\\\\v=\dfrac{-44+ \sqrt{44^2- 4(1)(-4000)} }{2(1)},\dfrac{-44- \sqrt{44^2- 4(1)(-4000)} }{2(1)}\\\\v=44.96\ ft/h, -88.96\ ft/h$

Since, 1 mph = 5280 ft/hour

44.96 ft/h = 0.00851 mph

-88.96 = -0.0168 mph

Hence, this is the required siolution.