Question

The braking distance (d) needed to stop a car varies directly with the square of the car's speed (s). A car going 30 mph needs 40 feet of braking distance to come to a complete stop. How many feet of braking distance would the same car, under the same conditions need going 70 mph? (Round the answer to the nearest whole number.)

Answer 1

This is the concept of proportionality;
Given that braking distance (d) varies directly with the speed then we shall have:
dαs^2
thus;
d=ks^2
where k is the constant of proportionality;
⇒k=d/s^2
when s=30mph, d=40ft
thus;
k=40/(30)^2=2/45
Hence our equation will be:
d=2/45s^2
thus the braking distance when the speed=70 mph will be:
d=2/45*(70)^2
d=217 7/9=217.78 feet

Answer 2

1. The braking distance is a function with variable speed.

2. So D(s)=
`k s^(2)`

which means the greater the speed (right side), the greater the distance needed to stop, which is natural.

k is a constant, like the weight of the car or other factors which are considered non changing in our case. So we are looking to identical cars, and the effect of the speed on the stopping distance.

3. D(30)=40=
`k (30)^(2)=900k`


`k= (40)/(900)= (4)/(90)`

4.
`D(70)=k* s^(2)= (4)/(90)* (70)^(2)= 218` feet