caro2789
Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?
A) 90
B) 130
C) 150
D) 320
E) 450
Solving for a distance "gap" in a "kiss" or "crash" scenario (traveling in opposite directions, toward one another).
At first the distance gap between them is 770 km. That number matters only once - it decreases when A travels alone.
Truck A travels
alone for one hour. RT=D. (90 kmh * 1 hr) = 90 km. Part of A's total distance is this 90 km.
Now the gap between them is 770 - 90 = 680 km
In a distance gap:
--if A and B travel in opposite directions, whether towards or away from one another, add their rates. (They close -- or expand -- the gap much faster when both move than when only one moves.)
--if vehicles travel the same direction (e.g. parallel train tracks, "chase"), subtract their rates
Add (or subtract) rates, and calculate time to meet,
only when both travel simultaneously.
Here, add rates: 90 + 80 = 170 kmh
\(\frac{Distance}{CombinedRate} = Time\) to meet
\(\frac{680}{170} = 4\) hours to meet
A's total distance:90 km alone.
Then (RT=D): (90*4) = 360 km
90 + 360 = 450 km
B 's total distance, RT = D
(80*4) = 320 km
A travels (450 - 320) = 130 km farther than B
Answer B