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 May 11th, 2015, 06:26 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Constant Acceleration A racing car is travelling at $130$ mph when the driver sees a broken-down car on the track $\frac{1}{10}$ of a mile ahead. Slamming the brakes on he achieves his maximum deceleration of $24.5$ mph per second. How far short of the broken-down car does he stop? My attempt, I know $u=130$, $a=-24.5$, which formula should I use to solve this question? May 11th, 2015, 04:07 PM   #2
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 Originally Posted by jiasyuen A racing car is travelling at $130$ mph when the driver sees a broken-down car on the track $\frac{1}{10}$ of a mile ahead. Slamming the brakes on he achieves his maximum deceleration of $24.5$ mph per second. How far short of the broken-down car does he stop? My attempt, I know $u=130$, $a=-24.5$, which formula should I use to solve this question?
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First convert a to use one unit for time (hour - using mixed makes things complicated).

Get time (t) from u-at=0. then get distance (s) from $\displaystyle s=ut-\frac{at^2}{2}$. May 12th, 2015, 04:53 AM #3 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 I can't get the answer after figuring very long time. The answer is 0.0042 miles. How to convert 24.5 mph per second to 25.5 miles/h^2? Last edited by jiasyuen; May 12th, 2015 at 05:04 AM. May 12th, 2015, 05:20 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Using mixed units can be confusing, but I think this one isn't too bad. It's a good rule of thumb to convert to SI when you begin. If you find that you've already performed a bunch of calculations and you don't want to redo everything, then go back over your calculations and check the units at each step. Below is an example where the units have been considered at every calculation step. initial speed = u = 130 mph final speed = v = 0 mph acceleration = a = -24.5 mph /s Let's get the time taken to stop: SUVAT equation: $\displaystyle v = u+at$ $\displaystyle t = \frac{v-u}{a} = \frac{0-130}{-24.5} = \frac{130}{24.5} = 5.306$ s This is in seconds because we have $\displaystyle [t] = \frac{mph}{(mph/s)}=\frac{mph \cdot s}{mph} = s$ Now we know the time taken to stop, we can get the distance travelled by the car as it's stopping: SUVAT equation: $\displaystyle s = ut + \frac{1}{2}at^2$ $\displaystyle = 130 \times 5.306 - \frac{24.5}{2}(5.306)^2$ $\displaystyle = 689.796 - 344.898$ $\displaystyle = 344.898$ but what are the units of this? We have $\displaystyle [s] = mph \cdot s = \frac{miles \cdot s}{hr}$ miles seconds per hour? Lovely. Let's convert the hours into seconds. We know that there are 3600 seconds in an hour, so $\displaystyle 1 \frac{miles \cdot s}{hr} = 1 \frac{miles \cdot s}{3600 s} = \frac{1}{3600} \frac{miles \cdot s}{s} = \frac{1}{3600} miles$ So $\displaystyle = 344.898 mph \cdot s = \frac{344.898}{3600} miles = 0.0958 miles$ So the distance, $\displaystyle d$, between the broken down car and the speeding car is: $\displaystyle d = 0.1 - 0.0958 = 0.0042 miles$ Thanks from jiasyuen Last edited by Benit13; May 12th, 2015 at 05:23 AM. May 12th, 2015, 05:52 AM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Benit. Thanks for your explanation. May 12th, 2015, 06:31 PM   #6
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 Originally Posted by jiasyuen I can't get the answer after figuring very long time. The answer is 0.0042 miles. How to convert 24.5 mph per second to 25.5 miles/h^2?
Multiply by sec/hr (3600). 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 hyperbola Physics 1 March 15th, 2015 10:10 PM 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 imcutenfresa Calculus 5 October 7th, 2009 02:51 PM

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