
#1
Aug509, 04:30 AM

P: 2

1. The problem statement, all variables and given/known data
A car is traveling at 80km/h, and brakes with a reaction time of .5 seconds. The car decelerates at 8m/s/s, determine the stopping distance of the car. 2. Relevant equations  V_{1} = V_{0} + at  D = V_{0}t + 1/2at^{2} Where: V_{1} = Final Speed V_{0} = Initial Speed A = Acceleration T = Time D = Distance 3. The attempt at a solution Distance Traveled before Braking: 80km/h = 22.22m/s 0.5 * 22.22m/s = 11.11 m Distance After Reaction: V_{1} = V_{0} + at 0 = 22.22  8t t = 2.7775 seconds D = V_{0}t + 1/2at^{2} D = 22.22 * 2.7775  [(8 * 2.7775^{2})/2] D = 30.86m Total Stopping Distance = 11.11m + 30.86m = 41.97 m Notes I am only beginning to learn some basic concepts of physics in class, so I'm not entirely sure about the equations. My teacher has provided no answer sheet to the question, and I am finding it hard to grasp the method of solving the question without a step by step example from a similar question. I'd greatly appreciate it if anyone could let me know if I am right, or how to do it if I am wrong. Thanks :D 



#2
Aug509, 05:11 AM

Mentor
P: 40,883

Looks good to meeverything's correct.
FYI, there's another kinematic equation that relates velocity and distance directly for accelerated motion. That would enable you to combine a few steps. See: Basic Equations of 1D Kinematics 



#3
Aug509, 06:57 AM

P: 2

Thank you for the help and additional formulas !
Now I'll be able to do my other sums :D 


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