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When a car's brakes are applied, it travels 5 feet less in each second than the previous second until it comes to a complete stop. A car goes 45 feet in the first second after the brakes are applied. How many feet does the car travel from the time the brakes are applied to the time the car stops?

higgsb  Sep 23, 2016
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Given a = -5 ft/sec^2

and xf = 45     and   t = 1        find vo

xf = xo + vot + 1/2 a t^2          xo = 0

45 = vo (1) + 1/2 (-5) 1^2       yields vo = 47.5 ft/s

 

Find TIME to stop  given  vo = 47.5 ft / s      a = -5 ft/s^2    and vf=0 (the car has stopped)

 

vf = vo + a t

0 = 47.5 - 5 t    yields  t = 9.5 s

 

Find DISTANCE to stop given  vo=47.5 ft/s  a = -5 ft/s^2   and  t= 9.5 s

xf=xo + vo t + 1/2 a t^2        xo=initial position being measured from = 0

xf = 47.5 (9.5) + 1/2 (-5)(9.5^2)    Yields   xf = 225.625 ft  to stop

ElectricPavlov  Sep 23, 2016

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