Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Uniformly accelerated motion under gravity

  1. May 18, 2016 #1
    Uniformly accelerated motion under gravity:- I have been taught that when a body accelerates, it travels on ratio of 1:3:5... so on. I have also been taught that at the last second, any body in the world will travel with the velocity of 5 m/s.
    Why is that so?
    Why I should believe on some axiom.
     
  2. jcsd
  3. May 18, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello Vivan, :welcome:
    This sounds very unbelievable to me. Can you specify where you picked this up and how exactly it was formulated ?
     
  4. May 18, 2016 #3
    It is written in many textbooks I have referred. Every where it is explained in the same way. It tells that though respective of how much the body travelled, it will travel with a velocity of 5 m/s in the last second.
    I don't understand why is that so....?
     
  5. May 18, 2016 #4
    You did not referred any. Give at least one example, with title, author, page. This is what it means to give a reference.

    The distances traveled during each of the first three seconds (for a body falling from rest) are indeed in the ratio 1:3:5. Their actual values are 5 m, 15m amd 25 m (taking g=10 m/s).

    The speed gained during each second, including the last second, is 10 m/s.

    But the speed "at the last second" is meaningless so it does not even make sense to argue if it's 5m/s or not.During the last second, as during any other second, the speed increases, by 10 m/s. So at the beginning of the last second the speed has one value and at the end another value.

    The closest you can come to your flawed statement may be that the average speed during the first second is 5m/s. Maybe you have a closer look at these textbooks.
     
    Last edited: May 18, 2016
  6. May 18, 2016 #5

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Since this is not true, I highly doubt that it is written in many textbooks. I, like nasu, would like to see the actual reference you are basing this on.
     
  7. May 19, 2016 #6
    Ok @nasu thanks for your help.
    I think that's the only way possible to understand.
    The book is Aakash module(Target 1).
    Well thanks for your explanation. I will think over it &I ask some more doubts.
     
  8. May 19, 2016 #7
    It seems it is a textbook used in India.
    Maybe you can post an image of the page where you think it says what you think it says. :)
     
  9. May 19, 2016 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you throw something straight up (assuming you throw it fast enough so that it travels upwards for more than a second), then:

    In the last second before it reaches its highest point it will travel ##5m## (hence have an average speed of ##5m/s## during this second); and, in the first second of its descent it will also travel ##5m##, with the same average speed during that second.

    Quite what the significance of this might be is anyone's guess! Really you should be thinking about motion under gravity as a velocity under a continuous rate of change; not as a sequence of discrete average speeds.
     
  10. May 19, 2016 #9
    Yeah but mainly UNIFORMLY ACCELERATED MOTION UNDER GRAVITY is a game of continuous rate of change working under gravity, Right?
    If that is true then I should conclude that motion under gravity is a relation of velocity with respect to time working under a constant(known as Gravity).
     
  11. May 19, 2016 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Gravity near the Earth's surface is an example of uniform acceleration. The point of kinematics is that, whatever the cause of the uniform accleration, the same kinematics principles, equations and solutions apply.

    What's special about gravity is that the acceleration is independent of the mass of the object, so that for all objects in freefall near the Earth's surface:

    ##\frac{dv_y}{dt} = g##

    Where ##v_y## is the vertical component of the velocity of an object. One reason the ##5m/s## "rule" makes no sense is that gravity does not affect the horizontal component of velocity, which remains constant during freefall.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Uniformly accelerated motion under gravity
  1. Motion under gravity (Replies: 5)

  2. Motion under gravity (Replies: 6)

Loading...