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Answer :
The percentage decrease in the braking distance of the car is 19.5%.
Distance traveled by the car
v² = u² + 2as
where;
- v is final velocity when brake is applied
- u is initial velocity
- s is distance traveled
s = (v² - u²)/(2a)
v = 100%u - 10.3%u = 89.7% u = 0.897u
s = [(0.897u)² - u²]/(2a)
s = -0.195u²/2a
percent decrease = 0.195 x 100% = 19.5%
Thus, the percentage decrease in the braking distance of the car is 19.5%.
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