At what time will the two buses be 274 miles apart?

Question

a bus leaves a station at 1 P.M., traveling west at an average rate of 44 m/hr. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/hr. At what time will the two buses be 274 miles apart?

Answer 1

2.5 hours

The car traveling west will travel 44 miles in the hour before the other car leaves. He will then travel at a rate of 44 mi/hr west. Distance = Speed * Time. Let's use the variable 't' to represent that time it takes for each car to get to the point where they're 274 miles apart. The car traveling west will travel (44t + 44) miles. Notice the 44 head start miles plus 44t, which is a speed of 44 * the time, giving you distance.
In the same amount of time, the other car will travel 48t (speed of 48 * the time). If you add those distances, it should come out to be 274. So let's setup the equation:

(44t + 44) + (48t) = 274

When you solve that equation for t, you get 2.5 hours. That's the time it took the car traveling east to reach its position, and the car traveling west to reach its position after it had already traveled 1 hour. So you add 1 hour to 2.5 hours to get 3.5 hours, which is the time it took from when the first car started moving.

I hope this helped!

Answer 2

At 2 pm the buses are 44 miles apart (the speed of the first bus times one hour). From then on they separate at a rate of 92 miles/hour (44 + 48 miles/hour). So at 3 pm they are 136 miles apart, at 4 pm they are 228 miles apart, and at 4:30 pm they are 274 miles apart.

You can think of it this way: from 1 to 2 they separate by 44 miles. From 2 to 3, and from 3 to 4, they separate by 92 miles. From 4 to 4:30, they separate by one-half of 92 miles, which is 46 miles. 44+92+92+46 = 274.

Answer 3

44(T - 1) + 48(T - 2) = 274
44T - 44 + 48T - 96 = 274
92T = 414
T = 4.5 ==> 4:30 PM