Hi i want to derive the 2nd equation of motion using the 1st

[faiziqb12]

Homework Statement

there are a lot of mathametical and graphical derivations of the three laws of motion but i have been trying to derive the second equation of motion from the first one but i always end hopeless.
please help

Homework Equations

1st equation v[f] = v + at
2nd equation s = vt + 1/2(at^2)

3. The Attempt at a Solution
i tried to use
v = d[f] / t[f]
v = d / t
in the first equation but it proved of no use

 

[Paradox101]

Hi
faiziqb12! Welcome to PF! I can't and won't give the answer directly, all I can do is guide you.
First, try recalling the equation of average velocity.Can any variable there be substituted from the first eqⁿ?

<Mod note: Removed excessive text formatting>

 

[DEvens]

Welcome to the forum.

How much calculus do you know? The way you do this depends on how much calculus you know.

It seems like the first equation should be

$v(t) = v_0 + a t$

That is, the first equation should give you velocity as a function of time, given an initial velocity and a constant acceleration. If that is right, then the second equation should be

$s(t) = s_0 + v_0 t + \frac{1}{2} a t^2$

which gives you the position as a function of time, given an initial position, an initial velocity, and a constant acceleration.

Can you show that the second equation follows from the first? Have you studied enough calculus for that? Do you know how to take a derivative?

 

[Paradox101]

DEvens, I don't think he knows about calculus.If however he did,all he would have to use would be d/dx(s/t)=v [or v=f'(s/t)] and ∫vtdt=s

 

[faiziqb12]

thanks all and especially paradox101
you showed me the way upto there!

but still i want it the way done in which i have done in the attempt as a solution section
please refer

 

[faiziqb12]

Welcome to the forum.

How much calculus do you know? The way you do this depends on how much calculus you know.

It seems like the first equation should be

$v(t) = v_0 + a t$

That is, the first equation should give you velocity as a function of time, given an initial velocity and a constant acceleration. If that is right, then the second equation should be

$s(t) = s_0 + v_0 t + \frac{1}{2} a t^2$

which gives you the position as a function of time, given an initial position, an initial velocity, and a constant acceleration.

Can you show that the second equation follows from the first? Have you studied enough calculus for that? Do you know how to take a derivative?

i don't know calculus
i am just a 9th standard whiz kid.
do you mean we can use the first equation standalone to get the second equation using calculus

 

[Zobrox]

Homework Statement

there are a lot of mathametical and graphical derivations of the three laws of motion but i have been trying to derive the second equation of motion from the first one but i always end hopeless.
please help

Homework Equations

1st equation v[f] = v + at
2nd equation s = vt + 1/2(at^2)

3. The Attempt at a Solution
i tried to use
v = d[f] / t[f]
v = d / t
in the first equation but it proved of no use

will post shortly... just editing
So first we know that velocity is the rate that distance changes, with respect to time:
$$v = \frac{dx}{dt}$$
if we rearrange the above, we get: (1) $$vdt = dx$$

Now... we can represent the change in distance as (2) $$dx = x-x_{0},$$ where$$x_{0}$$ is the initial distance.

now from the initial equation we know:

(4) $$v = v_{0} + a*\delta t$$

so using the equation above, we can substitute in (4) to (1):
however we have to compensate for velocity linearly increasing for the second term in (4)
so to do this we look at it in respect to dt/2:
therefore (1) becomes:

$$(v_{0} + \frac{1}{2}a*\delta t)(\delta t) = dx$$

so the final equation becomes:

$$dx = v_{0}*\delta t + \frac{1}{2}a*\delta t^2$$

However you would usually achieve this result through integration. Which in effect is finding the area under the curve, which is really the same as doing it graphically, however your using a formula rather than visual intuition.

 

[DEvens]

i don't know calculus
i am just a 9th standard whiz kid.
do you mean we can use the first equation standalone to get the second equation using calculus

Yes. Without calculus it is a bit harder, but it can still be done.

A graph of velocity vs time will have an interesting property. The area under the curve is the distance traveled up to that time. If you were at constant velocity it would just be a rectangle. That is $s(t) = v_0 t$. The graph is a horizontal line at height $v_0$. The area of a rectangle is just width times height.

But it's too easy.

A constant acceleration means that the velocity changes a constant amount in each second. So you draw yourself a graph. Across the bottom is the time. Up the left side is the velocity. You get a shape that is a rectangle with a triangle on top. The area under this is the distance traveled in any time.

So you have a rectangle that has a height of the initial velocity. And it is as wide as the time you are interested in. So the distance corresponding to the initial velocity is just $v_0 t$ because that is the area of that rectangle.

The triangle on top is again as wide as the time you are interested in. It is as high as $at$, the increase in velocity. So the area of this triangle is half it's width times its height. So $\frac{1}{2} a t^2$.

So you have the distance moved due to an initial velocity, and the extra distance moved due to a constant acceleration. But you started someplace. So your location is the second equation.

Try that with some actual graphs, and actual values of $a$.

Then relate that back to the average velocity. If you accelerate for time $t$ at acceleration $a$. You start at $v_0$ and finish at $v_0 + at$. The average speed is half the sum of these two. The distance traveled in time $t$ is just the average times the time. Or $v_0 t + \frac{1}{2} a t^2$. The same answer as before.

And always remember you started someplace. So you need $s_0$.

 

[PeroK]

An alternative suggestion is to draw a graph of velocity against time for constant acceleration. Displacement is then ...?

 

[faiziqb12]

th

Yes. Without calculus it is a bit harder, but it can still be done.

A graph of velocity vs time will have an interesting property. The area under the curve is the distance traveled up to that time. If you were at constant velocity it would just be a rectangle. That is $s(t) = v_0 t$. The graph is a horizontal line at height $v_0$. The area of a rectangle is just width times height.

But it's too easy.

A constant acceleration means that the velocity changes a constant amount in each second. So you draw yourself a graph. Across the bottom is the time. Up the left side is the velocity. You get a shape that is a rectangle with a triangle on top. The area under this is the distance traveled in any time.

So you have a rectangle that has a height of the initial velocity. And it is as wide as the time you are interested in. So the distance corresponding to the initial velocity is just $v_0 t$ because that is the area of that rectangle.

The triangle on top is again as wide as the time you are interested in. It is as high as $at$, the increase in velocity. So the area of this triangle is half it's width times its height. So $\frac{1}{2} a t^2$.

So you have the distance moved due to an initial velocity, and the extra distance moved due to a constant acceleration. But you started someplace. So your location is the second equation.

Try that with some actual graphs, and actual values of $a$.

Then relate that back to the average velocity. If you accelerate for time $t$ at acceleration $a$. You start at $v_0$ and finish at $v_0 + at$. The average speed is half the sum of these two. The distance traveled in time $t$ is just the average times the time. Or $v_0 t + \frac{1}{2} a t^2$. The same answer as before.

And always remember you started someplace. So you need $s_0$.

thats much like the graphical method..isnt it?
i don't think you derived it using calculus

 

[faiziqb12]

will post shortly... just editing
So first we know that velocity is the rate that distance changes, with respect to time:
$$v = \frac{dx}{dt}$$
if we rearrange the above, we get: (1) $$vdt = dx$$

Now... we can represent the change in distance as (2) $$dx = x-x_{0},$$ where$$x_{0}$$ is the initial distance.

now from the initial equation we know:

(4) $$v = v_{0} + a*\delta t$$

so using the equation above, we can substitute in (4) to (1):
however we have to compensate for velocity linearly increasing for the second term in (4)
so to do this we look at it in respect to dt/2:
therefore (1) becomes:

$$(v_{0} + \frac{1}{2}a*\delta t)(\delta t) = dx$$

so the final equation becomes:

$$dx = v_{0}*\delta t + \frac{1}{2}a*\delta t^2$$

However you would usually achieve this result through integration. Which in effect is finding the area under the curve, which is really the same as doing it graphically, however your using a formula rather than visual intuition.

thanks..I know integration somewhat.. i can use it ..

 

[Qwertywerty]

thats much like the graphical method..isnt it?
i don't think you derived it using calculus

Finding these formulae does not at all require calculus - the graphing method is completely fine .

Also , I'm confused . In a certain post you say you don't know calculus , but in another you say you know some integration .

So which is it ?

 

[faiziqb12]

e

Finding these formulae does not at all require calculus - the graphing method is completely fine .

Also , I'm confused . In a certain post you say you don't know calculus , but in another you say you know some integration .

So which is it ?

integration is just a small part of calculus
i know it because I am used with curves

 

[DEvens]

th

thats much like the graphical method..isnt it?
i don't think you derived it using calculus

At this level, the "graphical method" and calculus are pretty much the same thing.