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angelrenee2000
04.06.2021 •
Mathematics
The braking distance (dm) of a car is directly
proportional to the square of its speed (v m/s)
When the speed of the car is increased by
50%, what is the percentage change in its
braking distance?
Solved
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Ответ:
Step-by-step explanation:
We can assign variables as:
d = original breaking distance
b = original speed of the vehicle
d′ = increased breaking distance
b′ = increased speed of the vehicle
We can look at it this way, originally it was like this:
d=kb2
Now, if we increase the speed to b’ which 200% more than b, we will have
d′=k(b′)2
d′=k(b+2b)2
d′=k(3b)2
d′=9kb2
Obtaining the difference between d and d’, we have
d′−d=9kb2−kb2
d′−d=8kb2
in terms of d
d′−d=8(kb2)
d′−d=8d
Adding d on both sides of the equation,
d′=d+8d
Which means the original breaking distance is increased by 800%. This is similar to the phrase “the speed of the vehicle is increased by 200%” which translates to “b + 2b”.
Hope this answer helps you :)
Have a great day
Mark brainliest
Ответ: