SOLUTION: Tobias starts out jogging from Dauntless Headquarters at 6:00 a.m. at a speed of 4 mi/h. Twelve minutes later, Tris starts out from Dauntless Headquarters and follows the same rout
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Question 1001955: Tobias starts out jogging from Dauntless Headquarters at 6:00 a.m. at a speed of 4 mi/h. Twelve minutes later, Tris starts out from Dauntless Headquarters and follows the same route. At what constant rate in miles per hour must Tris run to catch Tobias at 6:42 a.m.? Put answer into nearest tenth decimal Found 2 solutions by josgarithmetic, MathTherapy:Answer by josgarithmetic(39618) (Show Source):
You can put this solution on YOUR website! RT=D; use time quantities instead of points on a time line; set up a data table for rates, times, distances. "Catch-up" is when Tobias and Tris went the same distance. Best unit for time will be in the equivalents as HOURS.
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Time QUANTITIES are used here in unit of HOURS or fractions of hours.
rate time in hours distance
Tobias 4 t+12/60 d
Tris r 42/60-12/60 d
The description is enough information to obtain two equations, but the system will have THREE unknown variables of r, t, and d.
--After some thought, t=42/60-12/60 and time becomes a knowable constant for Tobias and another for Tris; making t as known. Make this adjustment and the only unknowns are r and d.
You can put this solution on YOUR website!
Tobias starts out jogging from Dauntless Headquarters at 6:00 a.m. at a speed of 4 mi/h. Twelve minutes later, Tris starts out from Dauntless Headquarters and follows the same route. At what constant rate in miles per hour must Tris run to catch Tobias at 6:42 a.m.? Put answer into nearest tenth decimal
Distance Tobias has traveled in 42 (6:00 a.m. – 6:42 a.m.) minutes is: , or 2.8 miles
Time Tris takes to travel 2.8 miles = 30 (6:12 a.m. – 6:42 a.m.) minutes, or hour
Speed at which Tris needs to travel to catch Tobias at 6:42 a.m. = , or 2.8 * 2, or mph
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