Question 72383
Use the distance equation d=rt. You will have two distance equations
{{{d[1]=44t}}}The first bus heading west.
{{{d[2]=48(t-1)}}}The second bus heading east. (notice how the time is t-1. This is set up because for the first hour since the bus doesn't cover any distance). Now to find out how far each bus is away from each other, you simply add the two distances {{{d[1]}}} and {{{d[2]}}}. For instance, if one bus is 10 miles out and another is 5 miles out, then they are 10+5=15 miles away from each other.
{{{d[1]+d[2]=274}}}Set up equation
{{{(44t)+(48*(t-1))=274}}}Now solve for t
{{{44t+48t-48=274}}}
{{{92t-48=274}}}
{{{92t=274+48}}}
{{{92t=322}}}
{{{t=322/92}}}
{{{t=3.5}}}So it takes 3 and a half hours for the 2 buses to be 274 miles apart. To fully answer this question, 3 and a half hours after 1 PM is 4:30 PM.
Check:
{{{(44*3.5)+(48*(3.5-1))=274}}}
{{{154+120=274}}}
{{{274=274}}}
If that doesn't help, you can plug t=3.5 into the first equation d=44t (the first bus). You'll see that the first bus traveled 154 miles. For the 2nd bus, instead of the bus travelling 3.5 hours, it travels 2.5 (since it starts an hour later). So thats why you must use t-1 instead of t for the second bus. When you plug in this info, you'll see that the 2nd bus travels 120 miles. If you add these 2 distances, you get 274 miles, which shows that our answer works. Hope this helps.