Physics Physics Forum

 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: $\displaystyle d=d_r+d_a=\frac{475}{18}\,\text{m}+\frac{45125}{64 8}\text{ m}=\frac{62225}{648}\text{ m}\approx96\text{ m}$ Thanks from Benit13 Tags acceleration, constant Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Rollsroyce Algebra 1 November 16th, 2013 01:07 PM yogazen2013 Physics 1 September 25th, 2013 09:35 PM Mike7remblay Physics 3 February 1st, 2012 06:48 PM JosephE Algebra 2 February 13th, 2010 08:38 AM imcutenfresa Calculus 5 October 7th, 2009 02:51 PM

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