It's quicker than you think!
To stop a car, first you require a certain reaction time to begin breaking; then the car slows under the constant breaking deceleration. Suppose the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.5 km/h and 24.4 m when its initial speed is 48.3 km/h. Then your reaction time is (assuming reaction time is same for each case and breaks provide uniform retardation)
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1 solution
In the time equal to reaction time the car will move with constant velocity.
Let us assume reaction time to be ‘t’ and retardation of the car due to the breaks applied be ‘−a’.
Case I:
The total distance traveled by the car before coming to rest=56.7 m.
Hence 56.7=80.5(5/18) t + [ ( 80.5 *(5/18))^2 ]/2a -------(1)
Case II:
The total distance travelled is 24.4 m
Hence 24.4=48.3(5/18) t + [ ( 48.3 *(5/18))^2 ]/2a -------(1)
Solving (1) and (2) we get t=0.74 s Note: Above, multiplying the velocities with 5/18 is necessary to convert it into m/s