The distance s, in feet, a car needs to come to a stop on a certain surface depends on the velocity v, in feet per second, of the car when the brakes are applied. The equation is given by x=0.0344v^2-0.758v. What is the maximum velocity a car can have when the brakes are applied and stop within 150 ft?

Question

The distance s, in feet, a car needs to come to a stop on a certain surface depends on the velocity v, in feet per second, of the car when the brakes are applied.  The equation is given by x=0.0344v^2-0.758v.  What is the maximum velocity a car can have when the brakes are applied and stop within 150 ft?

anonymous · Answer

6.002233m/s

anonymous · Answer

how is it set up?  shouldn't the answer be in feet or am I misreading?

anonymous · Answer

i ve rechecked ....its d only possibility

amistre64 · Answer

$$s = 0.0344v^2 -0.758v$$
is this the equation to begin with?

anonymous · Answer

idk...this one confuses me.  This is why word problems are horrible - to much info that is not needed

amistre64 · Answer

$150 = 0.0344v^2 -0.758v$ ; now we -150 from both sides and solve for v

$0 = 0.0344v^2 -0.758v-150$  ; there is alot of decimals in this, but if you know your quadritic formula, it makes it bearable

anonymous · Answer

ok, i'm getting there. i'm breaking it down and solving for v

amistre64 · Answer

$$\frac{.758 \pm \sqrt{.758^2 -4(.0344)(-150)}}{2(.0344)}$$

amistre64 · Answer

when its all said and done; your velocity is about 78 feet per sec

anonymous · Answer

it comes out to like 77.9 so I can round up

anonymous · Answer

ur the best amistre!

amistre64 · Answer

:)  good luck