Brakes

The braking efficiency of a passenger car is required to stop at 12.5 m at an initial speed of 40 km/h. What is the acceleration braking by brakes?

Correct result:

a =  -4.938 m/s2

Solution:

v=40 km/hm/s=40/3.6 m/s=11.11111 m/s s=12.5 m v1=v+02=11.1111+025095.5556 m/s  t=s/v1=12.5/5.5556=94=2.25 s  a=0vt=011.11112.25=40081=4.938 m/s2v=40 \ km/h \rightarrow m/s=40 / 3.6 \ m/s=11.11111 \ m/s \ \\ s=12.5 \ \text{m} \ \\ v_{1}=\dfrac{ v+0 }{ 2 }=\dfrac{ 11.1111+0 }{ 2 } \doteq \dfrac{ 50 }{ 9 } \doteq 5.5556 \ \text{m/s} \ \\ \ \\ t=s/v_{1}=12.5/5.5556=\dfrac{ 9 }{ 4 }=2.25 \ \text{s} \ \\ \ \\ a=\dfrac{ 0-v }{ t }=\dfrac{ 0-11.1111 }{ 2.25 }=- \dfrac{ 400 }{ 81 }=-4.938 \ \text{m/s}^2



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