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Rate-Distance-Time word problems?
A cargo plane flew to the maintenance facility and back. It took one hour less time to get there than it did to get back. The average speed on the trip there was 220 mph. The average speed on the way back was 200 mph. How many hour did the trip there take?
The answer is 10, but I can't figure out how to get there!
A submarine left Hawaii tow hours before an aircraft carrier. The vessels traveled in opposite directions. The aircraft carrier traveled at 25 mph for nine hours. After this time the vessels were 280 mi. apart. find the submarine's speed.
Again, I have the answer, 5mph, but how do I figure it out?
Jose left the airport and traveled toward the mountains. Kayla left 2.1 hours later traveling 35 mph faster in an effort to catch up to him. After 1.2 hours Kayla finally caught up. Find Jose's average speed.
The answer is 20 mph.
Please help me to understand why!
1 Answer
- norcekriLv 71 decade agoFavorite Answer
(1) What is the time in each direction? (That's distance / rate, with distance as a variable). Write those two terms. Set their difference equal to 1 hour and solve for the distance.
This lets you get the two times as 10 and 11 hours.
(2) Let the submarine's speed be "s". it traveled for 11 hours. The carrier went 25 mph for 9 hours. The sum of those two is 280. Solve for "s".
(3) Let Jose's speed be "s" and his travel time is 2.1 + 1.2 hours.
Kayla travels at speed s+35 for 1.2 hours. Assuming that they left from the same place, they traveled the same distance. Set the two expressions equal to each other and solve for "s".
