neo007 Posted August 19, 2011 Share Posted August 19, 2011 (edited) Hi all, came across this question in prep for my GAMSAT exam. I've been trying to work it out, but my answers do not match up with the given answers in the book. 5.5 x 10^-7 is the given answer. I've attached the problem with this post. Can anyone verify that 5.5 x 10^-7 is correct or a mistake? Any help appreciated. edit: Ignore the graph part Edited August 19, 2011 by neo007 Link to comment Share on other sites More sharing options...
timo Posted August 19, 2011 Share Posted August 19, 2011 Any result without physical units is definitely wrong. 5.5 * 10^-7 parts of daylight time may be roughly correct, depending on season. Link to comment Share on other sites More sharing options...
DrRocket Posted August 19, 2011 Share Posted August 19, 2011 Hi all, came across this question in prep for my GAMSAT exam. I've been trying to work it out, but my answers do not match up with the given answers in the book. 5.5 x 10^-7 is the given answer. I've attached the problem with this post. Can anyone verify that 5.5 x 10^-7 is correct or a mistake? Any help appreciated. edit: Ignore the graph part Why don't you show us your reasoning and your answer, with appropriate units ? Link to comment Share on other sites More sharing options...
CraXshot Posted August 19, 2011 Share Posted August 19, 2011 Okay i'm also a student preparing for my exams, i used Newtons second law in standard form combined with in momentum and i found the contact time to differ from the anser in the book. Please tell me youre anser so I can correct or confirm it, i'm not allowed to just give it to you so sorry:( Link to comment Share on other sites More sharing options...
imatfaal Posted August 19, 2011 Share Posted August 19, 2011 Craxshot - not sure why you want to involve force and mass (ie N's 2 law) or momentum. Change of velocity over time period is average acceleration. The problem gives the initial velocity u and the final velocity v and the acceleration and you need to find the time elapsed t ; one of the basic equations of motion (SUVAT) that I learnt included all those terms. The standard equations of motion depend on a uniform acceleration - but you have been told the average acceleration, for an overall answer is there a difference? Neo - And Timo is quite right - you would lose a hatful of points for failing to include units Link to comment Share on other sites More sharing options...
CraXshot Posted August 19, 2011 Share Posted August 19, 2011 (edited) ok i did what I usually do and make thing too complicated, but the masses elliminated and in the end i just came to that basic eqaution with the same answer, so i stand incorrected, but yes what you said is a lot less complicated but the results are the same none the less, and I still don't get the anser you're book claims. Edited August 19, 2011 by CraXshot Link to comment Share on other sites More sharing options...
imatfaal Posted August 19, 2011 Share Posted August 19, 2011 Craxshot - oh yeah - how they managed to get an answer that is that small is beyond me. As Timo hinted it is about right in terms of the units of a day (but still by no mean correct - just order of magnitude correct). To know that an answer like that is wrong just use orders of magnitude to check it quickly - contact less than a millionth of a second, acceleration/change of speed thousands of m/s per second, change of speed tens of metres/second... Link to comment Share on other sites More sharing options...
DrRocket Posted August 19, 2011 Share Posted August 19, 2011 Okay i'm also a student preparing for my exams, i used Newtons second law in standard form combined with in momentum and i found the contact time to differ from the anser in the book. Please tell me youre anser so I can correct or confirm it, i'm not allowed to just give it to you so sorry:( Let's be clear here. You are the one asking for help. You can receive help by showng your rationale and work. We will not do your work for you. What you are "allowed" to do is your problem. If you want help from me, I require that you provide your work. I don't need the practice of solving freshman problems. Link to comment Share on other sites More sharing options...
timo Posted August 19, 2011 Share Posted August 19, 2011 (edited) @CraXshot: Let's be clear here. You are the one asking for help. You can receive help by showng your rationale and work. We will not do your work for you. What you are "allowed" to do is your problem. I think when asking "what is your result" CraXshot was addressing neo007 (who started this thread), not the general audience. And CraXshot is indeed not supposed to provide an answer for neo. Edited August 19, 2011 by timo 1 Link to comment Share on other sites More sharing options...
DrRocket Posted August 19, 2011 Share Posted August 19, 2011 I think when asking "what is your result" CraXshot was addressing neo007 (who started this thread), not the general audience. And CraXshot is indeed not supposed to provide an answer for neo. Good point. I stand corrected. Link to comment Share on other sites More sharing options...
swansont Posted August 19, 2011 Share Posted August 19, 2011 Hi all, came across this question in prep for my GAMSAT exam. I've been trying to work it out, but my answers do not match up with the given answers in the book. 5.5 x 10^-7 is the given answer. No, the given answer is not numerically right, assuming no change in units. Link to comment Share on other sites More sharing options...
techster Posted August 24, 2011 Share Posted August 24, 2011 (edited) You have been given the initial velocity and a final velocity. The thing here is that the final velocity vector is opposite in direction the the initial one. So, I can take them as 25 m/s and -20 m/s. Now, whats the change in velocity? Average acceleration (a) is given as 3000 m/s2. a=(dv/dt) you know 'a' and 'dv'. Find 'dt' I am an engineer and teaching is my passion. you can ask me questions at Techster . I'll be happy to help you out! Edited August 24, 2011 by techster Link to comment Share on other sites More sharing options...
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