Two cars leave the same city The first car is travelling at

Two cars leave the same city. The first car is travelling at 40 mi/hr. One hour later the second car leaves travelling at 55 mi/hr along the same road. In how many hours will the 2nd car overtake the first car?

Solution

Now we know that the 2nd car will have overtaken the first car when they have both travelled the same distance.
Let t=time travelled by the first car
then (t-1)=time travelled by the 2nd car (time to overtake)
distance(d)=rate(r) times time(t) or d=rt; t=d/r and r=d/t
Distance first car travels = 40*t
Distance 2nd car travels=55*(t-1)
When 2nd car overtakes, these distances are the same:
40*t=55(t-1) get rid of parens
40t=55t-55 subtract 55t from both sides
40t-55t=55t-55t-55 collect like terms
-15t=-55 divide both sides by -15
t=11/3 or 3.66667 hrs
t-1=3.66667-1=2.666667 hrs--------------time for 2nd car to overtake
CK
40*3.66667=55*2.66667
146.668=146.668