Car crash

On the road, with a maximum permitted speed of 60 km/h, there was a car crash. From the length of the vehicle's braking distance, which was 40 m, the police investigated whether the driver did not exceed that speed. What is the conclusion of the police, assuming a decelerated motion of the vehicle with a deceleration of 5 m/s2

Result

s =  27.778 m

Solution:

s1=40 m a=5 m/s2 v=60 km/h=60/3.6 m/s=16.66667 m/s  t=v/a=16.6667/51033.3333 s  s=12 a t2=12 5 3.33332250927.7778 m27.778 m  s1>s v>60s_{1}=40 \ \text{m} \ \\ a=5 \ \text{m/s}^2 \ \\ v=60 \ km/h=60 / 3.6 \ m/s=16.66667 \ m/s \ \\ \ \\ t=v/a=16.6667/5 \doteq \dfrac{ 10 }{ 3 } \doteq 3.3333 \ \text{s} \ \\ \ \\ s=\dfrac{ 1 }{ 2 } \cdot \ a \cdot \ t^2=\dfrac{ 1 }{ 2 } \cdot \ 5 \cdot \ 3.3333^2 \doteq \dfrac{ 250 }{ 9 } \doteq 27.7778 \ \text{m} \doteq 27.778 \ \text{m} \ \\ \ \\ s_{1}>s \ \\ v>60



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