<p>a train leaves the station at 11a.m. traveling at the rate of 40 miles per hour . a faster train leaves the same station at 1p.m. that afternoon and travels in the same direction on a parallel track at the rate of 60 miles per hour . at what time will the faster train overtake the slower one ?</p>
<p>answer 5p.m. </p>
<p>Let “t” be the time in hours it takes for the first train to be overtaken. Therefore the time the second train needs to overtake the first will be (t-2) since it departs two hours later. Since the distance traveled by each train is the same:
40t = 60 (t-2). Solve for t and get 6. So 6 hours after the first train departs will be 5PM.</p>
<p>Yould draw a quick line to help you visualize this problem. By the time the second train leaves, the first one is 2 x 40 or 80 miles away. The second one catches up at the rate of 20. Hence four more hours are needed and that means 5pm. </p>
<p>No real need for equations! </p>