Brakes of a car

For the brakes of a passenger car to be effective, it is prescribed that a car moving on a horizontal road at a speed of 40 km. A car must stop on the track 15.4 m. What is the deceleration of the car?

Correct result:

a =  4.008 m/s2

Solution:

v=40 km/hm/s=40/3.6 m/s=11.11111 m/s s=15.4 m  s=12at2  v=at s=12a(v/a)2  2 s=v2/a  a=v2/(2 s)=11.11112/(2 15.4)=4.008 m/s2v=40 \ km/h \rightarrow m/s=40 / 3.6 \ m/s=11.11111 \ m/s \ \\ s=15.4 \ \text{m} \ \\ \ \\ s=\dfrac{ 1 }{ 2 } a t^2 \ \\ \ \\ v=at \ \\ s=\dfrac{ 1 }{ 2 } a (v/a)^2 \ \\ \ \\ 2 \ s=v^2/a \ \\ \ \\ a=v^2 / (2 \cdot \ s)=11.1111^2 / (2 \cdot \ 15.4)=4.008 \ \text{m/s}^2



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