# Maths

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!!! :)))))))

1. 0
1. 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 %
or
new speed = 1600 percent of old speed
so
percent increase = 1600-100 = 1500% increase

posted by Damon

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