Question

The stopping distance, d ( in feet) for a truck moving at a velocity (speed) v miles per hour is modeled by the equation: d=0.04v^2+1.1v What is the speed of the truck if it takes 100 feet to stop? a. 12.5 mph c. 18 mph b. 38 mph d. 13 mph  

Answer 1

Its b-38 just took it

Answer 2

Answer:

b. 38 mph

Step-by-step explanation:

Here, the given equation that shows the stopping distance ( in feet ) when the truck is moving with the speed of v miles per hour,

$d=0.04v^2+1.1v$

If d = 100,

$100=0.04v^2+1.1v$

$\implies 0.04v^2+1.1v=100$

$0.04(v^2+\frac{1.1}{0.04})=100$

By adding the square of the half of the coefficient of v on both sides,

We get,

$0.04(v+13.75)^2=107.563$

$v+13.75=\pm \sqrt{107.563}$

$v=-13.75 \pm \sqrt{107.563}$

$\implies v=-13.75-\sqrt{107.563}\text{ or } v = 13.75+\sqrt{107.563}$

$\implies v\approx 38.1063\text{ or } v\approx -65.6063$

Since, the speed can not be negative,

Hence, the approximate speed of truck would be 38 mph.

Option 'b' is correct.