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November 9th, 2016, 09:26 PM  #1 
Newbie Joined: Aug 2015 From: Alberta Posts: 18 Thanks: 0  Linear Velocity & Acceleration Question
I'm having trouble with this question. A car traveling at a speed of 140km/h decelerates for 20 seconds. During this time it travels 500m. a) What is its deceleration? b) What is its speed after the deceleration? Answers in the book are a) 1.389 m/s^2, b) 40 km/h But I can't figure it out. 
November 9th, 2016, 10:02 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,097 Thanks: 972 
$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2$ You were given ... $\Delta x = 500 \, m$, $t=20 \, s$, and $v_0=140 \, km/hr$ You need to convert km/hr to m/s and solve for acceleration, which will be a negative value. Once you determine acceleration, you also have time and initial velocity. use the equation ... $v_f=v_0+at$ Final velocity will be in m/s ... you have to convert it to km/hr. 
November 23rd, 2016, 02:54 AM  #3  
Newbie Joined: Aug 2015 From: Alberta Posts: 18 Thanks: 0  Quote:
38.89 / 20 seconds = 1.94 m/s as my Deceleration.... which is off my a bit. Second questions what do you mean by at? Displacement / Time = Velocity so. 500 / 20 = 25 m/s Then I divide by 1000 to convert to km's as well as x by 3600 500 / 20 = 25 25 / 1000 = .025 .025 x 3600 = 90 km's So I'm off there as well Last edited by jayinwww; November 23rd, 2016 at 02:59 AM.  
November 23rd, 2016, 05:05 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,097 Thanks: 972 
You didn't use the equations I cited in my previous post ... fyi, $a \ne \dfrac{v_0}{t}$, it is $a=\dfrac{v_fv_0}{t}$. You can't use this equation to calculate acceleration because you do not have final velocity. $140 \, km/hr \cdot \dfrac{1000 \, m}{km} \cdot \dfrac{1 \, hr}{3600 \, sec} = \dfrac{350}{9} \, m/s$ $\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2 \implies a= \dfrac{2(\Delta x  v_0 \cdot t)}{t^2}$ $a= \dfrac{2 \left(500  \frac{350}{9} \cdot 20 \right) }{20^2} = \dfrac{25}{18} \, m/s^2$ Note $\dfrac{25}{18} \approx 1.389$ displacement/time is average velocity, not final velocity ... $v_f= v_0 + at$ ... this equation is a direct result of solving the equation $a=\dfrac{v_fv_0}{t}$ for $v_f$ $v_f = \dfrac{350}{9} + \left(\dfrac{25}{18}\right) \cdot 20 = \dfrac{100}{9} \, m/s$ $\dfrac{100}{9} \, m/s \cdot \dfrac{3600 \, sec}{1 \, hr} \cdot \dfrac{1 \, km}{1000 \, m} = 40 \, km/hr$ Final note ... if you are studying kinematics under uniform acceleration, you should have been exposed to these equations and the hints on when to use them. Last edited by skeeter; November 23rd, 2016 at 05:12 AM. 

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acceleration, linear, question, velocity 
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