Question 1047089

a car leaves at 7am on a long trip travelling at 44 kph a second car leaves at 11am traveling 108 kph what time does the second car catch the first
<pre>Let time 2nd (faster) car takes to get to meetup point, be T 
Then time 1st (slower) car takes to get to meetup point = T + 4
We then get the following <b><u>DISTANCE</b></u> equation: 108T = 44(T + 4)
108T = 44T + 176
108T - 44T = 176
64T = 176
T, or time 2nd (faster) car takes to get to meetup point = {{{176/64}}} = {{{highlight_green(matrix(1,5, 2&3/4, hours, or, 2.75, hours))}}}