Question 102852
Dorothy drives 10 kilometers, then increases her speed by 10 kilometers per hour and drives another 25 kilometers. Find her original speed if she drove for 45 minutes. 
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Let s = original speed
then
Increased speed = (s+10)
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Write a time equation: Time = distance/speed; Change 45 min to hrs: 45/60 = .75 hr
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Time for 10 km + time for 25 km = .75 hr
{{{10/s}}} + {{{25/((s+10))}}} = .75
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Multiply equation by s(s+10) and you have:
10(s+10) + 25s = .75(s(s+10)
:
10s + 100 + 25s = .75s^2 + 7.5s
:
35s + 100 = .75s^2 + 7.5s
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Arrange as a quadratic equation:
.75s^2 + 7.5s - 35s - 100 = 0
:
.75s^2 - 27.5s - 100 = 0
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Use the quadratic formula to solve for s: a=.75; b=-27.5; c=-100
{{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{s = (-(-27.5) +- sqrt( -27.5^2 - 4 * .75 * -100))/(2*.75) }}}
:
{{{s = (+27.5 +- sqrt(756.25 - (-300)))/(1.5) }}}
:
{{{s = (+27.5 +- sqrt(756.25 + 300))/(1.5) }}}
:
{{{s = (+27.5 +- sqrt(1056.25))/(1.5) }}}
:
{{{s = (27.5 + 32.5)/1.5}}}; we only want the positive solution here
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{{{s = 60/1.5}}}
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s = 40 mph is her speed for the 1st 10 km (original speed
:
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Check our solution using the time equation:
{{{10/40)}}} + {{{25/50}}} = .75 as given
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