Answer :
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%
Read more about distance
https://brainly.com/question/15172156
#SPJ1
The braking distance decreases by approximately 19.55% when the speed of a car is reduced by 10.3%, due to the kinetic energy being proportional to the square of the speed.
The question pertains to the relationship between the braking distance of a car and its speed. When the speed of a car is reduced, the kinetic energy of the car, which is proportional to the square of its speed, decreases. Since the braking distance is directly related to the kinetic energy that needs to be dissipated, a reduction in speed will lead to a greater percent decrease in braking distance than the percent decrease in speed.
To find the percent decrease in braking distance when the speed is reduced by 10.3%, we need to calculate the initial and final kinetic energies using the formula [tex]KE = 1/2 m v^2[/tex]. Assuming mass (m) remains constant, a 10.3% decrease in speed can be represented as multiplying the initial speed (v) by 0.897 (since 100% - 10.3% = 89.7%). Thus, we compare the squares of the initial and reduced speeds to find the percent decrease in kinetic energy, and consequently, the braking distance.
If we denote the initial speed as v and the final speed as [tex]v_f[/tex], we have [tex]v_f = 0.897v[/tex]. Thus, the new kinetic energy is [tex]KE_f = 1/2 m (0.897v)^2 = 0.8045 (1/2 m v^2)[/tex], which is approximately a 19.55% decrease in kinetic energy (since 100% - 80.45% = 19.55%). Therefore, the braking distance decreases by approximately 19.55%.
