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1/18
W/4 mph
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? A. 2 hoursB. 4 ½ hoursC. 5 ¾ hoursD. 6 hoursE. 7 ½ hours AnswerThe distance the 1st cyclist travels is 6(3+t) miles, andthe distance the 2nd cyclist travles is 10t miles...where t=the number of hours since cyclist #2 starts.Now solve: 6(3+t)=10t ==> 18+6t=10t ==> 18=4t ==> t=4.5 hrs==> 4.5 hours = 4 hrs 30 minutes
The average speed for the entire trip is 9mph.
1 and a half hours
1/18
W/4 mph
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? A. 2 hoursB. 4 ½ hoursC. 5 ¾ hoursD. 6 hoursE. 7 ½ hours AnswerThe distance the 1st cyclist travels is 6(3+t) miles, andthe distance the 2nd cyclist travles is 10t miles...where t=the number of hours since cyclist #2 starts.Now solve: 6(3+t)=10t ==> 18+6t=10t ==> 18=4t ==> t=4.5 hrs==> 4.5 hours = 4 hrs 30 minutes
Assuming you want to know how long before the second catches up with the first: After 3 hours the first cyclist has travelled 3 x 6 = 18 miles Second cyclist is travelling at 10mph, so in every hour he does 10-6 = 4 miles more. So to catch up the 18 miles between the two will take 18 / 4 = 4.5 hours = 4 hrs 30 mins
The average speed for the entire trip is 9mph.
The average speed is 8 and 8/9 mph.
1 and a half hours
Velocity = distance / time Given d = 72 miles ; t = 4 hour V = 72 miles / 4 hour = 18 miles per hour
9 mph and 18 mph
ANSWER # 1 It looks like cyclist #2 is out to catch cyclist #1. Our #2 cyclist is starting out 3 hours later than cyclist #1, and is traveling at 10 mph versus the 6 mph of cyclist #1. There are a couple of ways to solve this one, so there is no "right way" to do it. Opting for an easy one, let's think of it this way. Think of it like both cyclists are going the same speed, the 6 mph. Yes, our trailing cyclist can't ever catch the lead cyclist, but this is a short cut, okay? If both cyclists ride, the space between them neither increases or decreases. Here's the scoop. It's the space between them that we're gonna look at. That space is 3 hours long. Here's the thinking. Cyclist #2 is going faster than #1 by 4 mph, and that is the "extra" speed that our #2 cyclist is using to close the gap, to make up that 3 hours. If cyclist #2 is going to make up 3 hours at 4 mph (the differential speed), how much ground is he going to have to make up? Well, cyclist #1 is riding at 6 mph for 3 hours to create the gap. That's 6 time 3 or 18 miles that cyclist #2 is going to have to make up. And our #2 guy is going to have to make up the 18 miles at 4 mph (the differential speed). How long is it going to take him? The 18 miles divided by 4 mph equals 4 ½ hours. Presto. It will take our #2 cyclist that 4 ½ hours to catch up to cyclist #1. If we want to check our work, take cyclist #2's 10 mph speed and multiply by 4 ½ hours and we'll see that he has to ride 45 miles to catch cyclist #1. Let's see what happens for cyclist #1. This cyclist will be riding for 4 ½ hours plus the 3 hour head start. That's 7 ½ hours of riding at 6 mph. The 7 ½ times 6 equals 45 miles. Check! We're good here! ANSWER # 2 1st = 6(x +3) ......... 2nd = 10x 10x = 6(x+3) x = 4 (1/2)
An hour is 60 minutes, so the cyclist would travel 2.4 x 6=14.4 mph (miles per hour)
The distance is about 24 miles. A world-class cyclist might make it in a half-hour, while an average bicyclist would require about 2 hours at 12 mph.