Question

12. The braking distance of a vehicle is directly
proportional to the square of its speed. When the
speed of the vehicle is b m/s, its braking distance
is d m. If the speed of the vehicle is increased
by 200%, find the percentage increase in its
braking distance.​

Answer 1

Step-by-step explanation:

✮ Question :-

The braking distance of a vehicle is directly

proportional to the square of its speed. When the

speed of the vehicle is b m/s, its braking distance

is d m. If the speed of the vehicle is increased

by 200%, find the percentage increase in its

braking distance.

✮ Solution :-

➨An increase of 200% means the new speed is b + 2b = 3b.

➨Since the braking distance is proportional to the square of the speed, it is proportional to (3b)² = 9b²

Therefore,

$\: \: \: \: \: \\ {\bold{\sf{\boxed{\pink{Percent \: \: change \: \: = \: \: 100 × \: \: \frac{final \: - \: initial}{initial}}}}}}$

${\bold{\sf{➻ \: \: \: \: \: 100 \: × \: \frac{9b² - b²}{b²}}}} \\ </p><p>$

${\bold{\sf{➻ \: \: \: \: \: 100 \: × \: \frac{8b²}{b²}}}}$

${\bold{\sf{➻ \: \: \: \: \: 100 \: × \: 8}}}$

${\bold{\sf{\fbox{\orange{800 \% \: increase}}}}}$