motion problem

Suzi304

New member
Joined
Jan 26, 2006
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Hello, I need help with this problem. I can figure out the answer, but I cannot figure out how to set up the problem and solve. The question reads as this...

If the speed were increased by 10mph, a 420 mile trip would take 1 hour less time. How long will the trip take at the slower speed?


So far, I have

D=RT
420=x()
420=x+10()

What goes in the time part?

Please help
 
\(\displaystyle (r+10)(t-1)=420\)

\(\displaystyle rt=420--->r=\frac{420}{t}\)

\(\displaystyle (\frac{420}{t}+10)(t-1)=420\)

\(\displaystyle 420-\frac{420}{t}+10t-10=420\)

\(\displaystyle 410-\frac{420}{t}+10t=420\)

\(\displaystyle 10t^{2}+410t-420=420t\)

\(\displaystyle 10t^{2}-10t-420=0\)

Solve for t.

This is one of many ways to set this problem up.
 
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