when a driver brings a car to a stop by breaki.g as hard as possible the stopping can be regarded as the sum of reaction distance and breaking distance. tje following data are given, v1 m/s contains 10,20,30. reaction distance are 7.5, 15, 22.5. the breaking distance are 5.0, 20, 45. and thr last are 12.5, 35, 67.5. a) what is the reaction time around? b) what is the stopping distance if the velociry is 25 m/s?
need the ans. asap
The reaction time is t =x/v=7.5/10 = 15/20 =22.5/30 = 0.75 s.
The reaction distance for v=25 m/s is x=vt=25•0.75 = 18.75 m.
The acceleration is
a=v²/2s=10²/2•5=20²/2•20=30²/2•45=10 m/s²
The breaking distance for v=25 m/s is
s= v²/2a=25²/2•10=31.25 m
The stopping distance is x+s = 18.75+31.25=50 m
To find the answers to the given questions, we will follow the steps:
a) To determine the reaction time, we will first identify the pattern between the velocity and reaction distance:
Velocity (v1): 10 m/s, 20 m/s, 30 m/s
Reaction Distance: 7.5 m, 15 m, 22.5 m
From the given data, we can observe that the reaction distance increases linearly with the velocity. Thus, we can calculate the reaction time by dividing the reaction distance by the velocity:
Reaction Time = Reaction Distance / Velocity
For the given velocities, the reaction times would be:
10 m/s: 7.5 m / 10 m/s = 0.75 s
20 m/s: 15 m / 20 m/s = 0.75 s
30 m/s: 22.5 m / 30 m/s = 0.75 s
Therefore, the reaction time is 0.75 seconds.
b) To find the stopping distance when the velocity is 25 m/s, we need to sum the reaction distance and braking distance.
Velocity (v1): 25 m/s
Reaction Distance: 15 m
Braking Distance: 20 m
Stopping Distance = Reaction Distance + Braking Distance
Stopping Distance = 15 m + 20 m = 35 m
Therefore, the stopping distance when the velocity is 25 m/s is 35 meters.