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Mathematics 9 Online
OpenStudy (anonymous):

A thief is spotted by a policeman from a distance of 100 m. When the policeman starts the chase, the thief also starts running. If the speed of thief be 4 kmh" 1 and that of the policeman 10 kmh 1 , how far will the thief have to run before he is caught?

OpenStudy (aravindg):

why did you close i t?

OpenStudy (anonymous):

i want to post another question so i closed it.i will post it once again sorry

OpenStudy (aravindg):

ok

OpenStudy (anonymous):

A thief is spotted by a policeman from a distance of 100 m. When the policeman starts the chase, the thief also starts running. If the speed of thief be 4 kmh" 1 and that of the policeman 10 kmh 1 , how far will the thief have to run before he is caught?

OpenStudy (anonymous):

Ok this question is to do with relative speed, there are two main types of object relativity, one is absolute we use division and the other is relative we use addition/subtraction

OpenStudy (anonymous):

In your case the police is chasing the thief so they are both on the same direction, therefore the relative speed would be the P-T

OpenStudy (anonymous):

so 10km/hr - 4km.hr= 6km.hr

OpenStudy (anonymous):

the thief was spotted at 100m, hence 100m/6000m/hr=1/60hr, this means that 10Km/hr x 1/60hr=1/6 km

OpenStudy (anonymous):

which is about 166 m

OpenStudy (aravindg):

@mathsmind giving direct worked out answers is against openstudy code of conduct .Refer http://openstudy.com/code-of-conduct . Instead help the asker get to the solution being a guide along the process.

OpenStudy (callisto):

I think it is better to keep the comment relevant to the question. But.. I'm sorry that I am leaving some comments irrelevant to the question pooja123 posted. @mathsmind While we appreciate your intention to help, it is better to guide the user rather than doing all the work yourself. At least the asker can take part in the problem solving and hopefully, would learn something rather than just look at the last line and get the answer. (I am *not* referring to the asker of this question). If you think that this site started with a good and right intention, why not keep doing the right thing? @pooja123 It's okay, don't worry :) If anyone wants to continue with this discussion (about giving the entire solution), please post it in the Feedback group. Thank you!

OpenStudy (anonymous):

yes read all of my answers my friend, in all my posts, i go step by step, and i am not the only one who does that, so why me among all?

OpenStudy (anonymous):

read the step by step guide through solution please @Callisto

OpenStudy (amistre64):

This site has people in place to encourage the users to conduct themselves in a "proper" manner. If the askers do not participate in the solution process, then there is no point in continuing to hand over the answers. This is not a free answering service, it is a study site to help promote growth and learning through real time interactions. Any static webpage can give a step by step result that can be googled. Lets try to use the strengths of this site for growth and learning :)

OpenStudy (aravindg):

^^best said :)

OpenStudy (anonymous):

yes my friend read the step by step answer and judge with justice not with hearing

OpenStudy (amistre64):

A thief is spotted by a policeman from a distance of 100 meters. When the policeman starts the chase, the thief also starts running. (so they both start at the same time) )If the speed of thief be 4 kmh and that of the policeman 10 kmh, how far will the thief have to run before he is caught? when the time is the same is my first idea

OpenStudy (amistre64):

can you tell me what you think a good equation or relationship would be to solve this?

OpenStudy (anonymous):

i went through the process

OpenStudy (anonymous):

read my answer it up there i am not going to resolve it

OpenStudy (anonymous):

it's*

OpenStudy (anonymous):

i told u its relative speed

OpenStudy (amistre64):

this is an example of how you elicit a response and engage the asker ... if they are willing to participate, then you carry on ....

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