Asked by Anonymous
An employee drove to work on Monday at 45 mi/h and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mi/h and arriving one minute late.
a. How far does the employee live from work?
b. At what speed should the employee drive to arrive five minutes early?
a. How far does the employee live from work?
b. At what speed should the employee drive to arrive five minutes early?
Answers
Answered by
tchrwill
Using V = D/T
60D/45 = T - 1
60D/40 = T + 1
Solving for T and equating yields
60(40D - 45D)/1800 = 2 or 5D = 60 making D = 12 miles
In order to arrive in 17 minutes,
60(12)/V = 17 yielding = 42.35mph.
60D/45 = T - 1
60D/40 = T + 1
Solving for T and equating yields
60(40D - 45D)/1800 = 2 or 5D = 60 making D = 12 miles
In order to arrive in 17 minutes,
60(12)/V = 17 yielding = 42.35mph.
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