SOLUTION: a car leaves san diego at 1 pm traveling at a constant rate of 40 mph toward LA. 30 mins later, another car leaves san diego traveling toward LA at a constant speed of 55 mph. Afte
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: a car leaves san diego at 1 pm traveling at a constant rate of 40 mph toward LA. 30 mins later, another car leaves san diego traveling toward LA at a constant speed of 55 mph. Afte
Log On
Question 500895: a car leaves san diego at 1 pm traveling at a constant rate of 40 mph toward LA. 30 mins later, another car leaves san diego traveling toward LA at a constant speed of 55 mph. After what time is the distance between the cars 15 miles? Answer by geetha_rama(94) (Show Source):
You can put this solution on YOUR website! Let Speed of Car A = 40mph (which leaves at 1:00PM)
Let Speed of Car B = 55mph (which leave at 1:30PM)
Distance the car A would have covered in 30 mins = 40 * 30/60 = 20 miles.
At 1:30, Car A would be 20 miles from San diego and Car B at San diego.
Assume after time t, the distance between the two cars is 15 miles
Distance covered in time t by Car A = 40t + 20miles (distance covered from 1 to 1:30 PM)
Distance covered in time t by Car B = 55t
40t + 20 - 55t = 15 miles (distance between both cars is 15 miles)
-15t = -5
t = 1/3 hrs
= 1/3 * 60 = 20 mins
Hence the time is 1:30 + 20 mins = 1:50 PM
At 1:50PM the distance between two cars will be 15 miles
CHECK:: Distance covered by Car A (which starts at time 1:00PM )
= 40*(1/2+1/3) = 40*5/6 = 200/6 = 100/3 miles
Distance covered by car B (starting at 1:30PM) = 55*1/3 = 55/3 miles
Difference in the distance = 100/3 - 55/3 = 45/3 = 15 miles
(