T O P I C R E V I E W 
kanta 
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
1Sophia walks 10 minutes reaching 10(6)/60 = 1km from the starting point. 2Danish starts out at 15km/hr 1km away from Sophia 3Danish closes in on Sophia at the rate of (156) = 9km/hr. 4Danish reaches Sophia in t = 1/9 = .111..hr or 6min40sec. 5Danish 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. 

