March 15th, 2015, 09:54 PM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Constant acceleration
Determine stopping distances for a car with initial speed of 95km/h and human reaction time 1.0s for an acceleration a) $\displaystyle a = 5.0ms^{2}$ I worked through it and got an answer of 135m The answer is 94m. Need some help please. I start my physics paper in a few months so trying to get a head start Last edited by hyperbola; March 15th, 2015 at 09:59 PM. 
March 15th, 2015, 10:10 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
I would first convert the initial speed to m/s: $\displaystyle v_i=95\frac{\text{km}}{\text{hr}} \cdot\frac{1000\text{ m}}{1\text{ km}} \cdot\frac{1\text{ hr}}{3600\text{ s}}=\frac{475}{18}\,\frac{\text{m}}{\text{s}}$ So, we know that during the 1 second reaction time, the care traveled: $\displaystyle d_r=\frac{475}{18}\,\text{m}$ Now, for a constant acceleration of $a$ (in meters per second per second), we may use: $\displaystyle d_a=\frac{v_i^2}{2a}$ So, for part a), we are given: $\displaystyle a=5\frac{\text{m}}{\text{s}^2}$ Hence: $\displaystyle d_a=\frac{\left(\frac{475}{18}\,\frac{\text{m}}{\text{ s}}\right)^2}{2\left(5\frac{\text{m}}{\text{s}^2}\right)}=\frac{45125}{ 648}\text{ m}$ And thus the total stopping distance is: $\displaystyle d=d_r+d_a=\frac{475}{18}\,\text{m}+\frac{45125}{64 8}\text{ m}=\frac{62225}{648}\text{ m}\approx96\text{ m}$ 

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acceleration, constant 
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