Posted - 04/09/2012 : 14:06:31 Sophia sets off on a walk at a speed of 6 km/h.ten minutes later,her brother Danish sets off after her on his bicycle at 15 km/h. How far must Danish go to catch up with sophia?

4 L A T E S T R E P L I E S (Newest First)

TchrWill

Posted - 04/23/2012 : 09:58:38 Originally posted by kanta[/i]

Sophia sets off on a walk at a speed of 6 km/h.ten minutes later,her brother Danish sets off after her on his bicycle at 15 km/h. How far must Danish go to catch up with Sophia?

Alternatively

1--Sophia walks 10 minutes reaching 10(6)/60 = 1km from the starting point. 2--Danish starts out at 15km/hr 1km away from Sophia 3--Danish closes in on Sophia at the rate of (15-6) = 9km/hr. 4--Danish reaches Sophia in t = 1/9 = .111..hr or 6min-40sec. 5--Danish overtakes Sophia after riding d = 15(1/9) = 1.666...km.

kanta

Posted - 04/21/2012 : 04:28:24 hello dear.The time is given 10 minutes and you have solved the problem with 15 minutes.

kanta

Posted - 04/19/2012 : 06:44:51

quote:Originally posted by the_hill1962

Since you only stated the problem without commenting on what is giving you trouble, then we wonder if you are just looking for an answer given to you. Maybe you don't understand anything about the problem? In either case, please respond to the following hint and we will help you further: The equation that relates distance, rate and time is d=rt Maybe you already knew this. In the type of problem you are asking about, you have two objects and you want to know when their distances will match up (be equal). You set the two d=rt equations equal to each other and solve. Here is an example: Object1 going 7mph and object2 leaving 15 minutes later going 20mph. You want to know when (or where) the 20mph catches up to the first object that left. Well, the distance equation for object1 is d=7t and the distance equation for object2 is d=20(t-.25). Now, many students may ask about where and what the "t-.25" means. If you understand it, then great! If not, ask us. So, the answer to this example would be to solve 7t=20(t-.25) and that is t=5/13 (hours) That is about 23 minutes. If you want the distance, then put t=5/13 into either d=7t OR d=20(t-.25) Both should give you the same answer since you set them equal to get that value of 5/13. By the way, for this example, it came out to be d=2+9/13 miles.

Thank you

the_hill1962

Posted - 04/10/2012 : 11:31:03 Since you only stated the problem without commenting on what is giving you trouble, then we wonder if you are just looking for an answer given to you. Maybe you don't understand anything about the problem? In either case, please respond to the following hint and we will help you further: The equation that relates distance, rate and time is d=rt Maybe you already knew this. In the type of problem you are asking about, you have two objects and you want to know when their distances will match up (be equal). You set the two d=rt equations equal to each other and solve. Here is an example: Object1 going 7mph and object2 leaving 15 minutes later going 20mph. You want to know when (or where) the 20mph catches up to the first object that left. Well, the distance equation for object1 is d=7t and the distance equation for object2 is d=20(t-.25). Now, many students may ask about where and what the "t-.25" means. If you understand it, then great! If not, ask us. So, the answer to this example would be to solve 7t=20(t-.25) and that is t=5/13 (hours) That is about 23 minutes. If you want the distance, then put t=5/13 into either d=7t OR d=20(t-.25) Both should give you the same answer since you set them equal to get that value of 5/13. By the way, for this example, it came out to be d=2+9/13 miles.