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Thread: Objects travelling at different speeds and one starts travelling later

  1. #1
    New Member markl77's Avatar
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    Objects travelling at different speeds and one starts travelling later

    Okay so I need a kind of way to do these specific questions

    Object A travels at a constant velocity of 2.0m/s east, object B travels in the same direction at a constant velocity of 3.0 m/s. If object B starts 1 minute after object A, how long will it take object B to catch object A?

    I tried to start by like averaging them? but I don't even remember where to start / how to do these questions at all.

    anyways, I tried finding the avg velocity which is 5.0m/s , then I think I found the average displacement by using the time (60s) multiplied by the avg velocity to get 300m...
    then I tried just finding the displacement of object A which I found to be 120m. If B starts 60s after A, the question is asking for how long it would take for B's displacement to equal A's displacement ...

    I tried by using the formula d=vt for B's time at 120 metres to see where they would meet but I got 40s
    120m=3.0m/s(t)
    40s=t

    that's the wrong answer..
    Last edited by markl77; 02-05-2018 at 10:18 PM.
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    Quote Originally Posted by markl77 View Post
    Okay so I need a kind of way to do these specific questions

    Object A travels at a constant velocity of 2.0m/s east, object B travels in the same direction at a constant velocity of 3.0 m/s. If object B starts 1 minute after object A, how long will it take object B to catch object A?

    I tried to start by like averaging them? but I don't even remember where to start / how to do these questions at all.

    anyways, I tried finding the avg velocity which is 5.0m/s , then I think I found the average displacement by using the time (60s) multiplied by the avg velocity to get 300m...
    then I tried just finding the displacement of object A which I found to be 120m. If B starts 60s after A, the question is asking for how long it would take for B's displacement to equal A's displacement ...

    I tried by using the formula d=vt for B's time at 120 metres to see where they would meet but I got 40s
    120m=3.0m/s(t)
    40s=t

    that's the wrong answer..
    There are many ways to solve a problem like this. Guessing that averaging will do it is not one of them. Do you have a specific reason to think it would help? If so, then that would be worth considering. (By the way, the average of 2 and 3 is not 5, anyway.)

    One way is to think about how far apart they are when B starts, and how fast the distance between them is decreasing.

    Another is to write an expression for the distance between them at time t seconds after B starts, and use that to write an equation you can solve (which will say, as you suggested, that the displacement of each of them after 60 seconds is equal). Using this algebraic method, a key is to clearly define what the variable, and any expression you write, means. (Your equation says that after 60 seconds, B will be where A was when B started, which is not quite what you want.)

    If this is for a course you are taking, or from a textbook you are reading, can you find an example of a somewhat similar problem, and show us what method they used? You don't necessarily have to copy their method to solve it, but they probably are expecting you to be learning such a method, so you might as well do so!

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    New Member markl77's Avatar
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    Quote Originally Posted by Dr.Peterson View Post
    There are many ways to solve a problem like this. Guessing that averaging will do it is not one of them. Do you have a specific reason to think it would help? If so, then that would be worth considering. (By the way, the average of 2 and 3 is not 5, anyway.)

    One way is to think about how far apart they are when B starts, and how fast the distance between them is decreasing.

    Another is to write an expression for the distance between them at time t seconds after B starts, and use that to write an equation you can solve (which will say, as you suggested, that the displacement of each of them after 60 seconds is equal). Using this algebraic method, a key is to clearly define what the variable, and any expression you write, means. (Your equation says that after 60 seconds, B will be where A was when B started, which is not quite what you want.)

    If this is for a course you are taking, or from a textbook you are reading, can you find an example of a somewhat similar problem, and show us what method they used? You don't necessarily have to copy their method to solve it, but they probably are expecting you to be learning such a method, so you might as well do so!
    Yeah I was thinking of something else I guess, because in my notes it says to find average velocity you divide total displacement by total time so I guess that I assumed that adding them up includes this? I know that to actually find the average is to add them up and then divide by 2(or the number of velocities). Anyways, the work that is in the key from my textbook just says :
    let d(a)= displacement of a
    d(b)= displacement of b

    d(a)=d(b)
    120m+d(a)=d(b)
    120m+(3.0m/s)t=(2.0m/s)t
    120m=1.0m/s(t)
    120s=t

    what im confused about is how they came up with displacement of a=displacement of b and then they just added 120m to it..
    But ill try again using what you said about the expression
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    Quote Originally Posted by markl77 View Post
    d(a)=d(b)
    120m+d(a)=d(b)
    d(a)=d(b) doesn't make sense. 120m+d(a)=d(b) does. Do you understand where it comes from?

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    Draw representive diagram at time when B starts;
    at that time, A has travelled 2m:

    *****(2)*****(A)@2m/s.........................>?

    (B)@3m/s...............................................>?
    I'm just an imagination of your figment !

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    Quote Originally Posted by lev888 View Post
    d(a)=d(b) doesn't make sense. 120m+d(a)=d(b) does. Do you understand where it comes from?
    Actually, both do make sense, but with different definitions of d(a). I hope the text didn't really say both! One of the main things they should be teaching is the importance of defining variables and sticking to them. A couple other things must have been copied wrong.

    When they meet, both have traveled the same distance from the starting point (but in different times). What they should have said is something like this:

    let d(a)= displacement of a
    d(b)= displacement of b

    d(a)=d(b)
    va*1+vat=vbt
    [that is, the total displacement of a is distance in the first 60 sec plus distance in t sec]
    120m+(2.0m/s)t=(3.0m/s)t
    120m=1.0m/s(t)
    120s=t

    Also, I rarely include units in an equation while I solve it; that just gets in the way. But I often do include it when I first write the equation, in order to be sure units match up. So I would be solving the equation

    120 + 2.0t = 3.0t

    As we have mentioned, there are several other ways you work; this is in between the two ways I suggested.

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    Quote Originally Posted by Dr.Peterson View Post
    Actually, both do make sense, but with different definitions of d(a).
    Yes, this is a more precise way to describe the 2 equations

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    (distance behind) / (difference in speeds) also works...right?
    I'm just an imagination of your figment !

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    Quote Originally Posted by Denis View Post
    (distance behind) / (difference in speeds) also works...right?
    Yes, that's what I had in mind in my first suggestion:

    Quote Originally Posted by Dr.Peterson View Post
    One way is to think about how far apart they are when B starts, and how fast the distance between them is decreasing.
    That tends to be a less "algebraic" approach, so many courses will not mention it, but I like showing a variety of approaches, even though it demonstrates that algebra is not (always) necessary. My second suggestion was the traditional formal approach; the book as shown separately works out the "distance behind" before writing the equation.

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