Respuesta :
recall your d = rt, distance = rate * time.
let's say by the time the car gets in the highway, the truck has already been running for 5 minutes, and say by the time they meet the truck has been running for "t" hours, so the car has then been runnning for 5 minutes less than "t", now, since "t" is hours well, 5 minutes is just 5/60 or 1/12 of "t", so the car has been running when they meet for "t - 1/2"
now, just before the car passes the truck, they first have to meet, at that point, the distance travelled by both is exactly the same, say "d" miles.
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Truck&d&60&t\\ Car&d&74&t-\frac{1}{12} \end{array} \\\\\\ \begin{cases} \boxed{d}=60t\\ d=74\left( t-\frac{1}{12} \right)\\ ----------\\ \boxed{60t}=74\left( t-\frac{1}{12} \right) \end{cases} \\\\\\ 60t=74t-\cfrac{37}{6}\implies \cfrac{37}{6}=14t\implies \cfrac{37}{84}=t[/tex]
which is about 26 minutes and 25 seconds.
let's say by the time the car gets in the highway, the truck has already been running for 5 minutes, and say by the time they meet the truck has been running for "t" hours, so the car has then been runnning for 5 minutes less than "t", now, since "t" is hours well, 5 minutes is just 5/60 or 1/12 of "t", so the car has been running when they meet for "t - 1/2"
now, just before the car passes the truck, they first have to meet, at that point, the distance travelled by both is exactly the same, say "d" miles.
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Truck&d&60&t\\ Car&d&74&t-\frac{1}{12} \end{array} \\\\\\ \begin{cases} \boxed{d}=60t\\ d=74\left( t-\frac{1}{12} \right)\\ ----------\\ \boxed{60t}=74\left( t-\frac{1}{12} \right) \end{cases} \\\\\\ 60t=74t-\cfrac{37}{6}\implies \cfrac{37}{6}=14t\implies \cfrac{37}{84}=t[/tex]
which is about 26 minutes and 25 seconds.
