posted by StuartKess .
The braking distance of a car is directly proportional to te square of it's speed.
When the speed is p metres per second, the braking distance is 6m. When the speed is increased by 300%, find
(a) an expression for speed of the car
(b) the braking distance
(c) the % increase in the braking distance
Is there any fixed methods for this type of question? Thanks a lot!!! :)))))))
d = k v^2 so v = sqrt(d/k)
6 = k p^2 so p = sqrt (6/k)
new speed = 4 p = 4 sqrt (6/k)
new d = k v^2 = k*16(6/k) = 96 meters
96/6 = 16 times = 1600 %
new speed = 1600 percent of old speed
percent increase = 1600-100 = 1500% increase