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I have been working through the book called "Mathematics" written by A.D. Aleksandrov, A.n. Kolmogorov and M.A. Lavrent'ev recently and have had some difficulty with understanding Examples given by the authors regarding Mathematical Analysis which is the main topic in Chapter 2.

I understand the process of working out the increment $\Delta s=s_2-s_1=\dfrac{g}{2}(2t\Delta t+\Delta t^2)$ which represents the distance covered in the time from $t$ to $t + tΔ$.

The authors now describe the process of finding the average velocity over the section of path Δs by dividing Δs (which we know from the equation above) by Δt. This is then shown in the equation below.

$v_{av}=\dfrac{\Delta s}{\Delta t}=gt+\dfrac{g}{2}\Delta t$ This is the part of the Example which I don't understand. Firstly what is the value of Δt and how does one know what to divide by? And how to the authors get a velocity of gt + g/2 x Δt? Is there a 'value' for Δt and if yes how does one know it? Have I missed anything?

I have a link for the ebook version 'Mathematics'. The example which is involved in my question can be found on page 66 and 67 (Example 1) This shows the full example and can be read for further understanding when answering my question..... See here.

I hope this helps and Thank you for any answers in advance.

xAly
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  • They are just using the definition average velocity = (distance covered in some interval of time) / (time interval) here. $\Delta t$ is some positive quantity, which you (eventually) make arbitrarily small. – user_of_math Dec 23 '14 at 15:05
  • So this would mean that vav=Δs/Δt=gt+g/2xΔt is a 'fixed' equation/ value? – xAly Dec 24 '14 at 07:27

1 Answers1

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$s_1$ is the distance covered in time t and $s_2$ is the distance covered in time$(t+\Delta t)$.

$s_1=\dfrac{gt^2}{2}$ , and $s_2=\dfrac{g(t+\Delta t)^2}{2}=\dfrac{gt^2}{2}+\dfrac{g}{2}(2t\Delta t+\Delta t^2)$

Now, distance covered in the time interval $\Delta t$ is $\Delta s=s_2-s_1=\dfrac{g}{2}(2t\Delta t+\Delta t^2)$

The average speed in the time interval $\Delta t$ is $v_{av}=\dfrac{\Delta s}{\Delta t}=gt+\dfrac{g}{2}\Delta t$

And letting $\Delta t$ approach $0$ can give us exact velocity at that point. So

$v= \displaystyle\lim_{\Delta t \to 0}v_{av}=\displaystyle\lim_{\Delta t \to 0}gt+\dfrac{g}{2}\Delta t=gt$

Dheeraj Kumar
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  • How do you work out average speed without knowing Δt? More specifically, how do you get to the point vav=Δs/Δt=gt+g/2 x Δt? – xAly Dec 24 '14 at 07:28
  • for finding average speed we need to have the distance covered in that time interval and if we don't know the time $\Delta t$ our speed will be in terms of t, which you can clearly see it is – Dheeraj Kumar Dec 24 '14 at 07:43
  • What do you mean by 'in terms of t'? – xAly Dec 24 '14 at 11:37
  • "in terms of" means that the velocity is dependent on $\Delta t$ and we don't know its value. We just used it to find the velocity after time $t$ as you can clearly see on next page that we put $\Delta t \rightarrow0$ – Dheeraj Kumar Dec 24 '14 at 12:43
  • 'approach' zero means that whatever value for Δt gets smaller and smaller , but can never equal zero because that would lead to a math error right? – xAly Dec 24 '14 at 13:31
  • if $\Delta t$ becomes really small then we can easily neglect it – Dheeraj Kumar Dec 24 '14 at 13:35
  • @LostAce: It's not a math error but that it is impossible to divide something by $0$, so $\frac{Δs}{Δt}$ is invalid if $Δt = 0$. Also, we define velocity to be what that ratio approaches as $Δt$ gets smaller and smaller (whether negative or positive). Velocity may not always exist given the path taken. For example if a car can instantaneously start moving at $1m/s$, its velocity at the point it starts moving is undefined because at that point the ratio is $0$ when $Δt < 0$ and $1$ when $Δt > 0$. – user21820 Jan 01 '15 at 11:15