Krz Posted February 28, 2006 Posted February 28, 2006 well my physics class was assigned some homework the other day, and not even the teacher knew how to do this question (i live in a very small town ) anyway if anyone can help me out with this : "An unmarked police car, traveling a constant 90km/h, is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the police man steps on the accelerator. If the police car's acceleration is 2.00 m/s^2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at a constant speed)?" Umm I just noticed there's a homework forum, eheh sorry
The Thing Posted February 28, 2006 Posted February 28, 2006 If I'm not incorrect, I think I got an answer of 1.588 seconds. Could someone else check on that? Someone that actually KNOWS Physics? (lol not me, I used all math)
Krz Posted February 28, 2006 Author Posted February 28, 2006 the answers 15.8, i guess i should have posted that earlier. I just dunno how to get it
timo Posted February 28, 2006 Posted February 28, 2006 the answers 15.8, i guess i should have posted that earlier. I just dunno how to get it By using the ansatz Position_Speeder(time) = Position_Police(time) and solving for time.
swansont Posted February 28, 2006 Posted February 28, 2006 the answers 15.8, i guess i should have posted that earlier. I just dunno how to get it What equations apply here? As Athiest notes, you want to have an equation that will let you solve for the position of each car, so you can set them equal.
swansont Posted February 28, 2006 Posted February 28, 2006 Is this supposed to be a running gag? No, just mangled typing. Transposing i and e doesn't jump out at me like other typos do when I proofread. I assume the same when I'm call swansot, swasnont, or some such. Don't attribute to malice that which can be explained by incompetence.
s pepperchin Posted March 1, 2006 Posted March 1, 2006 I solved this problem the way Atheist said. I used the equations: Position_Speeder(time) = [math](13.88 m)+(38.88 \frac{m}{s}) t[/math] Position_Police(time) = [math](25 \frac{m}{s}) t+\frac{1}{2}(2.00 \frac{m}{s^2})t^2[/math] these equations are set up so that t=0 is when the police man begins accelerating, so after you set these equal to each other and solve for t you need to add one second for the second it took the police man to react.
Klaynos Posted March 1, 2006 Posted March 1, 2006 I would sugest that the first equation is for velocity not position
s pepperchin Posted March 1, 2006 Posted March 1, 2006 No, it is for position. Look at the units they all work and if you work out the problem you get the right answer. I would show you how I got it but I think that everyone should try it without seeing the process before I post the steps I used.
The Thing Posted March 2, 2006 Posted March 2, 2006 Oh frick... I know what I did wrong... Didn't convert from km to meters... I'm pretty sure my equation's right: [math] 25x+x^2-\frac{350}{9}x=\frac{350}{9}-25 [/math]
Krz Posted March 2, 2006 Author Posted March 2, 2006 thanks for the replies but what is this "ansatz Position_Speeder(time) = Position_Police(time)"? is it a well known formula for physics? i havent even heard of it before, i just saw the question solved using factoring and whatnot :S but you guys made it look way easier ^^
swansont Posted March 2, 2006 Posted March 2, 2006 "ansatz" is German term; used in science it means a made-up equation for a specific situation (vs a standard equation from a theory that can be applied generally)
s pepperchin Posted March 3, 2006 Posted March 3, 2006 The answer is about 14.8257 you forgot the second it takes for the police car to start accelerating.
Cloud Posted March 3, 2006 Posted March 3, 2006 acceleration = change in velocity / time 2.00 m/s2 = + 50 kph (0.014 kilometres per second) / ? 2/0.014 = 142.857 + 1 second = 143.857 seconds -> 2.4 minutes _____________________________________________________________ Oh damn - I just realised the actual speed has to be above 140. I'm not that good at physics calculations anyway. I always think too one-dimensionally.
s pepperchin Posted March 3, 2006 Posted March 3, 2006 acceleration = change in velocity / time 2.00 m/s2 = + 50 kph (0.014 kilometres per second) / ? 2/0.014 = 142.857 + 1 second = 143.857 seconds -> 2.4 minutes _____________________________________________________________ Oh damn - I just realised the actual speed has to be above 140. I'm not that good at physics calculations anyway. I always think too one-dimensionally. You are off by a factor of 100 somewhere.
Cloud Posted March 3, 2006 Posted March 3, 2006 What equations apply here? As Athiest notes, you want to have an equation that will let you solve for the position of each car, so you can set them equal. Yeah - there should be two equations which can be solved simultaneously.
Cloud Posted March 3, 2006 Posted March 3, 2006 You are off by a factor of 100 somewhere. Yes - well they don't teach this advanced stuff in high school. This is more towards primary college level material. Am I right?
s pepperchin Posted March 3, 2006 Posted March 3, 2006 This problem is not outside the scope of high school physics. If you haven't been shown how to do this in high school than your teacher is falling down on the job.
EvoN1020v Posted March 3, 2006 Posted March 3, 2006 From the given 2 constant speed from the question, you can use the graphic calculator to find the intersection point. Be careful: One of the equations should be y= nx + 1, where n is the value of the constant, and 1 is the passing second when the police accelerates. When you find the intersect point, find the distance of the line, and you should have the answer. Sorry if I sound confusing.
timo Posted March 4, 2006 Posted March 4, 2006 I think the question has already been beaten to death, resurrected and twisted into some weird abnomination (it´s only one equation to solve @Cluod and there is only one velocity that remains constant over time @Evon). It would be nice to have some feedback from Krz to know if he found the answer he was looking for (ideally with a description of what he understood and how he solved it) or if there´s some additional questions. Afterwards, there´s absolutely no problem if someone who knows how to solve the problem presents the full solution (approach, solving the solutions and plugging in the numbers - and please don´t supress the units!).
Star Caribou Posted March 4, 2006 Posted March 4, 2006 you forgot the second it takes for the police car to start accelerating. So the answer must be somewhere around 15.8 seconds.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now