The percent of the braking distance is mathematically given as
P=91.5128%
What percent does the braking distance of a car decrease, when the speed of the car is reduced by 10.3 percent?
Generally, the equation for is mathematically given as
V^2=U^2+2as
Therefore
V^2=u^2+2(-a)s
[tex]s=\frac{u^2}{-2a}[/tex]
Hence.
u'=u-u×0.103
u'=0.206u
v^2=u^2+2as
0=(0.206)^2+2(-a)s'
Hence
[tex]s'=\frac{0.042436u^2}{a}[/tex]
In conclusion,The reduction percentage is
[tex]P=\frac{s-s'}{s}*100\\\\P=\frac{(\frac{u^2}{-2a})-(\frac{0.042436u^2}{a} )}{ \frac{u^}{-2a}}*100\\\\P=\frac{(0.5-0.042436 )}{ 0.5}*100[/tex]
P=91.5128%
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