Problem Solving: Find the time for the second bus to catch up
A bus leaves a station at 8:00 a.m. and averages 30 mi/h. Another bus leaves the same station following the same route two hours after the first and averages 50 mi/h. When will the second bus catch up to the first?
Answer STEP 1:
We are given that the first bus leaves a station at a rate of 30 mi/h at 8.00 am and that the second bus leaves the same station two hours later at a rate of 50mi/h. Assume a variable for the time taken by the first bus to travel, say, t
Recall the formula:
In this case, the distances are the same, so we just need to set the rate × time
equal for both the cases.STEP 3:
We can see that 50 is distributed over the difference t – 2
. To remove the parentheses
in this expression, first multiply 50
by t – 2
. To remove the parentheses in this expression, first multiply t
. We get 50t
. Multiply 50
next. The result that you get is 30t = 50t – 100
Now, we have to isolate the variable terms on one side and the constant terms on the other side.
from both sides of the obtained expression and simplify
Isolate the variable t
, by dividing both sides of the obtained expression by -20
This is the time of travel for the first bus till the second bus catches up with it. STEP 6:
Now you have to find the time at which the second bus catch up to the first bus. We know that the first bus leaves the station at 8 00 am. So we have to add 5 hrs to the starting time of the first bus:
8.00 am + 5 hours = 1.00 pm
The second bus catches up to the first by 1.00 pm.