HARE and TOM chase
look at this..
TOM stands exactly 250m due south from a HARE
HARE starts moving towards EAST with a velocity Vm/s
TOM chases the HARE in such a way that..
at each instant he moves in the direction of HARE..
in other words he always heads towards the HARE
at that instant
now we have to find the distance travelled by both ,TOM and HARE when they meet
Amit
Monday, July 07, 2003
Consider relative velocity of HARE with respect to TOM.Find the time taken(T) for them to meet.Distance travelled by TOM would be V*T.Distance travelled by HARE would be tricky to calculate though.
Arun Iyer
Wednesday, July 09, 2003
i dont think the approach of relative velocity would solve the purpose..since at each instant the direction of the velocity of Tom is changing ..so you wont get a straight line path for TOM's travel..instead u will get a curve
and at each point ,the tangent to the curve will give the direction of velocity of tom which will meet the point where the hare will be at that instant.
Now how do u form an equation which gives u a general formula for Tom's velocity ?
amit
Friday, July 11, 2003
As you say the tangent gives the direction of velocity.Now what you do is resolve HARE's velocity along the direction of TOM's velocity and calculate the "instantaneous" relative velocity of "approach".Integrate this over time.Now this integral is nothing but equal to 250.(Can you see why?a figure should give you a reasonable idea)
This should give you a equation with a unknown i.e the angle made by the tangent with horizontal.To eliminate this we need another equation.For this we simply resolve TOM's velocity along HARE's velocity(i.e horizontal).Note that at the end of time T they both have covered equal horizontal displacements.
You simply solve these two equation to get your answer for Time T.Once T is found then HARE's distance can be found.(i made a silly error above and interchanged names :) ).
However finding the distance travelled by TOM seems to be a hurdle.But if it were displacement then Pythagoras theorem gives the answer.
Arun Iyer
Saturday, July 12, 2003
great..
now i can think on it.
as u say the equation which we need to form for integration is a bit tricky.
anyway if we can find HARE's distance then everything is done.
because whatever distance hare travels, that multiplied by 4/3 will give us the distance by TOM ..
got it?[we have been given tom travels at a speed 4/3 times that of hare ..so done..
amit
Monday, July 14, 2003
"tom travels at a speed 4/3 times that of hare" ?? did i miss it in the question or is it some trick??
Arun Iyer
Monday, July 14, 2003
I'm interested in this problem but I'm completely baffled by what has been posted so far. Would someone care to clean up the problem statement and write it in properly capitalized and punctuated English, using complete sentences, and not leaving out any vital information (e.g. "4/3")?
anon
Monday, July 14, 2003
Rephrasing the entire question ...
1.TOM and HARE are standing 250m apart.
2.HARE starts moving with velocity V in the direction perpendicular to the distance between them.
3.At the same time ,TOM starts moving with a velocity U(say) such that he is always pointing towards HARE.
4.Now find the distance that would have been travelled by both TOM and HARE before they meet.
hopefully,this is understandable
Arun Iyer
Thursday, July 17, 2003
i don't know anything about the 4/3 part though.i think that should be cleared up by the questioneer.
arun
Arun Iyer
Friday, July 18, 2003
sorry guys..
Take the speed of Hare as V and that for TOM as 4V/3
can anyone tell me what the answer i and how they could solve it..
amit
Sunday, July 20, 2003
If i have not made any calculation error or any silly mistake then i get,
HARE's distance = 3000/7 units
TOM's distance = 4000/7 units
Arun Iyer
Sunday, July 20, 2003
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