Exercises - Uniformly Accelerated Motion
Exercise 1:
A climate activist stuck himself on the street in the city center. A car approaching at 50 km/h brakes immediately. What braking distance does the car need if a reaction time of one second has to be taken into account? The braking deceleration (negative acceleration) is -8 m/s2.
The reaction distance sreact represents a uniform, straight-line movement:
sreact = v0 ∙ treact = 50 km/h ∙ 1s = 13,89 m/s ∙ 1s = 13,89 m
( 50km/h = 50 000 m / 3600 s = 13,89 m/s )
The braking distance sbraking can be determined by a negatively accelerated movement:
To use this equation, you must know that the braking distance must be calculated at the point in time at which the velocity has reached the value 0 m/s. We started with a speed of 13.89 m/s. Thus follows:
v2 - v02 = (0 m/s )2 - (13,89 m/s)2 = - (13,89 m/s)2
The total stopping distance is the sum of the reaction distance and the braking distance:
stotal = sreact + sbraking = 13,89 m + 12,06 m = 25,95 m