Question

Car X leaves Northtown traveling at a steady rate of 55 mph. Car Y leaves 1 hour later following Car X, traveling at a steady rate of 60 mph. Which equation can be used to determine how long after Car X leaves Car Y will catch up?

Answer 1

I need help on this question

Answer 2

the velocities are 55mph and 60mph, but car Y leaves 1 hour later, so you set an euqation like this
\[\huge{55(h)=60(h-1)}\]
solvinf for h you will find h=12, so 12. in 12 hours car X travels 55(12)=660
for the car Y you calculate 12-1 because he leaves 1 hour later,
so 60(12-1)=660, same distance

Answer 3

Another way to solve this (perhaps to check your work, as it doesn't supply the equation the problem asks for) is to observe that Car X will travel 55 miles during its 1 hour head start. Car Y travels 60 mph and Car X travels 55 mph, so Car Y closes Car X's lead by 60-55 = 5 miles each hour. How many hours will it take to overcome that 55 mile lead, if each hour whittles it down by 5 miles?

Answer 4

Note also that @lucaz equation will give you the total number of hours Car X has been driving, and the problem asks for a closely related but different quantity: how long after Car X leaves will Car Y catch up.