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Homework Help: Question about doubling speed of object

  1. Sep 3, 2012 #1
    I know that something with double the speed will travel 4 times as far. So does that mean you take the given distance and multiply it by 4? or take the distance and multiply it by 2 and square it? for example if its 3 meters would it be 3*4 or (2x3)^2
  2. jcsd
  3. Sep 3, 2012 #2
    This statement is meaningless because other conditions are not specified.
  4. Sep 3, 2012 #3
    Let me provide more information. The problem is:

    a woman walks back to her campsite, 2.16 km away. Her dog runs ahead, at twice the speed, reaches the campground and turns around until it meets her again. Then, the dog proceeds to run back to the campground again. This continues until the woman reaches her campground. What distance did the dog run?
    (use km as units)

    So, being that the dog is traveling twice as fast wouldnt I have to take 2.16km and assume that it is 4 times as much... However I dont know if its 4(2.16)or (2x2.16)^2 OR would I have to first assume the dog has already doubled the distance of 2.16 to 4.32 and then quadruple that??? I know it may seem like such a simple problem but I am stuck!
  5. Sep 3, 2012 #4
    You have constant speeds and you have equal time. Something moving with speed v over time t will cover distance d = vt. Something moving with speed 2v over time t will cover distance d = 2vt.
  6. Sep 3, 2012 #5
    This isn't close to how you begin to solve this problem. Let 2.16 km be represented by the symbol d. Let x represent the distance of the woman from the campsite at time t, where t = 0 corresponds to the time at which she and the dog were both at the distance d. Let v = the speed of the woman, and 2v represent the speed of the dog. The first step is to express x as a function of d, v, and t. Let y represent the distance of the dog from the campsite at time t, the first time that he is running ahead of her toward the campsite. Express y as a function of d, v, and t. This is just the start of the solution. There are additional steps after these.
  7. Sep 3, 2012 #6
    So I would have 2.16=vt and 2.16=2vt but I dont have velocity for either one or time except for the initial time of 0...I don't feel any closer to solving the problem. We havent gone over anything like this in class so I'm very confused :(
  8. Sep 3, 2012 #7


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    Given that 2.16 = vt, then what distance would 2vt be?
  9. Sep 3, 2012 #8
    4.32? How can the dog have only traveled double when in his first trip to the campsite and back he would already have covered that amount?
  10. Sep 3, 2012 #9


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    So you first said the woman was walking, now you are saying she is standing still, having never moved (else, how could the dog have traveled that amount in one round trip?)
  11. Sep 3, 2012 #10
    lol, no she's walking but since the question says that the dog goes and comes back and continues to do so...I imagine that he covers a lot of ground...but yes, I guess if she is walking then he actually wouldnt make it back to the starting point at all.

    I guess I'm confused because we have talked about how an object traveling twice as fast will travel 4 times as far so I keep thinking something like 4(2.16km) or (2X2.16)^2

    So basically it is d=2vt which is 4.32km I guess I was just complicating things :/
  12. Sep 3, 2012 #11


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    Yeah, that's a pretty good guess.

    You do realize, I assume, that this is a nonsensical statement?
  13. Sep 3, 2012 #12
    Sorry, my prof. has been going over how an object moving at twice the speed of that of another will travel 4 times as far (what I said before didnt make sense....but now that I look back at my notes it says that the "stopping" distance will be 4 times as far...which is a different concept altogether...:sigh:....I hope I don't mess this up on my exam

    I really do appreciate everyone's help, thanks...Physics is hard :/
  14. Sep 4, 2012 #13
    It is not really that hard. The woman needs time t to cover distance d with speed v. So t = d/v. All this time the dog is roving about with speed 2v, so its distance is (2v)t = (2v)(d/v) = 2d.
  15. Sep 4, 2012 #14
    After solving this problem by a complicated method (taking into account the detailed movement of the dog and woman) and seeing the final answer, I realized that this is somewhat of a trick problem. Voko recognized the trick immediately. The dog's path is irrelevant to this problem as long as he travels the same amount of time as the woman and arrives at the same location (the campsite). He is traveling twice as fast as the woman, so, in the same amount of time, he travels twice as far. You must have misinterpreted your professor regarding traveling 4X as far. The 4X factor applies to constant acceleration and deceleration. The present problem is a straightforward rate-time-distance problem.
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