given
vi = 65 miles/hour
= 65*1.609*5/18 m/s
= 29.05 m/s
vf = 0
a = -4.25 m/s^2
reaction time, delta_t = 0.3 seconds.
distance travelled during reaction time, d1 = vi*t
= 29.05*0.3
= 8.715 m
distance traveled during break applied, d2 = (vf^2 - vi^2)/(2*a)
= (0^2 - 29.05^2)/(2*(-4.25))
= 99.28 m
stoping distance = d1 + d2
= 8.715 + 99.28
= 108 m <<<<<<<<<---------------------Answer
Determine the stopping distance, in meters, for an automobile traveling with an initial (constant) speed of...
PART A: Determine the stopping distances for an automobile going a constant initial speed of 95 km/h and human reaction time of 0.30 s , for an acceleration a=−3.6m/s2. Express your answer using two significant figures. PART B: Determine the stopping distances for an automobile going a constant initial speed of 95 km/h and human reaction time of 0.30 s , for an acceleration a=−6.5m/s2. Express your answer using two significant figures.
Determine the stopping distance for an automobile with an
initial speed of 95km and human reaction time of 1s for an
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a car is traveling at a constant speed of Vo on a straight, level road. After the brakes are applied at t=0 seconds, the motion can be approximated by: x=t^3/100-t^2 +17t where x is the distance traveled in meters and t is the time in seconds. Determine: A the initial velocity (Vo), B. the time required for the car to stop, C. The distance required for the car to stop, and D. The maximum deceleration during braking.