# Is this equation wrong?

yu_wing_sin
Is this equation wrong?

This is a equation for
"uniform acceleration directed line motion at zero initial speed" in our textbook.

S=1/2 at^2

Here is the list to compare the answer:
t: 1 2 3
v: 2 4 6

If we use the equation to calculate, the answer is different to the list.

For example,

1/2(2)(3*3) = 9, s=9

but in the list, it shows the answer is 12!

Is my wrong? Or the equation is really wrong?

Last edited:

## Answers and Replies

Staff Emeritus
yu_wing_sin said:
Is this equation wrong?

This is a equation for
"uniform acceleration directed line motion at zero initial speed" in our textbook.

S=1/2 at^2

Here is the list to compare the answer:
t: 1 2 3
a: 2 4 6

If we use the equation to calculate, the answer is different to the list.

For example,

1/2(2)(3*3) = 9, s=9

but in the list, it shows the answer is 12!

Is my wrong? Or the equation is really wrong?

It is wrong because the kinematical equation that is used to derived that expression is for constant acceleration only. If the acceleration varies, that equation isn't valid.

Zz.

yu_wing_sin
ZapperZ said:
It is wrong because the kinematical equation that is used to derived that expression is for constant acceleration only. If the acceleration varies, that equation isn't valid.

Zz.

But in my trial calculation, the a is already constant. The a is 2, it is still not equal to the list result. Is me there is any wrong?

Staff Emeritus
yu_wing_sin said:
But in my trial calculation, the a is already constant. The a is 2, it is still not equal to the list result. Is me there is any wrong?

What list? All you did was plug in t and a, and calculate s. YOu got an value for s. Why is the eqn. wrong?

Zz.

edit: I see you EDITED your list to now list v. This is no longer relevant and appropriate for the equation you are using. How could you plug "a" for "v"? This is getting VERY confusing.

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yu_wing_sin
ZapperZ said:
What list? All you did was plug in t and a, and calculate s. YOu got an value for s. Why is the eqn. wrong?

Zz.

edit: I see you EDITED your list to now list v. This is no longer relevant and appropriate for the equation you are using. How could you plug "a" for "v"? This is getting VERY confusing.

The reason of a comes from v is because the v is a speed with initial zero speed, so the first second, 1/s' speed is the a. But anyway I calculate it, I really can't get the true result.

Staff Emeritus
yu_wing_sin said:
The reason of a comes from v is because the v is a speed with initial zero speed, so the first second, 1/s' speed is the a. But anyway I calculate it, I really can't get the true result.

And you don't see ANYTHING wrong with plugging in the values of "v" into "a" in

$$s = \frac{1}{2} at^2$$?

If there is an INITIAL velocity at t=0, then your question is MISSING a term. The full equation for the displacement is

$$s= s_0 + v_0t + \frac{1}{2} at^2$$

It is ONLY for s(t=0)=0 and v(t=0)=0 do you get the displacement equation that you quoted in the first place.

Zz.

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yu_wing_sin
More details,

t: 1 2 3
v: 2 4 6

The v is a speed with zero initial speed, front mentioned. So the first time unit is the a, thus the a is 2.
If we put 2 into a for calculating, we will get:

1/2 a t^2
=1/2 (2) 3^2
=9

but the list shows that 3 seconds (t) = 12 (v). This is really making me doubt, whether the equation is wrong.

Staff Emeritus
yu_wing_sin said:
More details,

t: 1 2 3
v: 2 4 6

The v is a speed with zero initial speed, front mentioned. So the first time unit is the a, thus the a is 2.

What the......?

If we put 2 into a for calculating, we will get:

1/2 a t^2
=1/2 (2) 3^2
=9

but the list shows that 3 seconds (t) = 12 (v). This is really making me doubt, whether the equation is wrong.

I give up. Maybe someone else can translate what's going on.

Zz.

yu_wing_sin
ZapperZ said:
And you don't see ANYTHING wrong with plugging in the values of "v" into "a" in

$$s = 1/2 at^2$$?

If there is an INITIAL velocity at t=0, then your question is MISSING a term. The full equation for the displacement is

$$s= s_0 + v_0t + 1/2 at^2$$

It is ONLY for s(t=0)=0 and v(t=0)=0 do you get the displacement equation that you quoted in the first place.

Zz.

How come is the s_0 and v_0?
The initial velocity is already the a, and the v is variable, and the s_0 where can find?

Let me trial calculate it, ignore the s_0,

v_0t+1/2 at^2
=2*3 + 1/2 (2) (3^2)
=15

only "v_0t+1/2 at^2" result is ready far higher than this equation "S=1/2 at^2".

Gold Member
The equation s=1/2at^2 goves the total distance traveled. After 3 seconds this is 1/2*2*3^2=9. The list says that after 3 seconds the velocity is 6. Where is the problem with this?

Gold Member
yu_wing_sin said:
How come is the s_0 and v_0?
The initial velocity is already the a, and the v is variable, and the s_0 where can find?

Let me trial calculate it, ignore the s_0,

v_0t+1/2 at^2
=2*3 + 1/2 (2) (3^2)
=15

only "v_0t+1/2 at^2" result is ready far higher than this equation "S=1/2 at^2".
Why are you taking v_0 = 2? If the accelration is constant then you can extrapolate v_0 to be 0.

Nylex
If you're starting at postition s = 0 and time t = 0, the equation is just

$$s = v_{0}t + \frac{1}{2}at^2$$

If you're starting with velocity v = 0 at time t = 0, the equation reduces further

$$s = \frac{1}{2}at^2$$

The $s_{0}$ and $v_{0}$ are just the displacement and velocity at time t = 0, respectively. What's hard to understand? ZZ's explanation seemed pretty clear to me :/.

yu_wing_sin
LeonhardEuler said:
The equation s=1/2at^2 goves the total distance traveled. After 3 seconds this is 1/2*2*3^2=9. The list says that after 3 seconds the velocity is 6. Where is the problem with this?
This is an accelerated motion, the a is 2 , so the volecity is not wrong.

v=at
v=2*3
v=6

Staff Emeritus
yu_wing_sin said:
How come is the s_0 and v_0?
The initial velocity is already the a, and the v is variable, and the s_0 where can find?

I'm sorry, but INITIAL VELOCITY is "a"? Since when can an acceleration becomes a velocity? Where did you learn this?

Zz.

yu_wing_sin
LeonhardEuler said:
Why are you taking v_0 = 2? If the accelration is constant then you can extrapolate v_0 to be 0.

Oh, sorry. I missed understanding to it. In my textbook there is no this mention.

yu_wing_sin
LeonhardEuler said:
Why are you taking v_0 = 2? If the accelration is constant then you can extrapolate v_0 to be 0.

Oh, I missed understanding to it. In my textbook there is no mention of this. But the result is still wrong.

Gold Member
ZapperZ said:
I'm sorry, but INITIAL VELOCITY is "a"? Since when can an acceleration becomes a velocity? Where did you learn this?

Zz.
He seems to have gotten this from the fact that it is a uniform acceleration, so:
$$a = \frac{\Delta v}{\Delta t} = \frac{v_1-v_0}{1-0}=2$$

Gold Member
yu_wing_sin said:
Oh, I missed understanding to it. In my textbook there is no mention of this. But the result is still wrong.
Is it possible that the initial position is not zero? Could you explain to me exactly how you know that the result is wrong. I don't seem to see any contradiction within the data you gave.

yu_wing_sin
ZapperZ said:
I'm sorry, but INITIAL VELOCITY is "a"? Since when can an acceleration becomes a velocity? Where did you learn this?

Zz.

The initial speed is at zero, that is not wrong. It has many motive examples, from still to accelerate. It is also used for calculating certain time's distance. But I found it is not correct.

Staff Emeritus
LeonhardEuler said:
He seems to have gotten this from the fact that it is a uniform acceleration, so:
$$a = \frac{\Delta v}{\Delta t} = \frac{v_1-v_0}{1-0}=2$$

Yes, but he's going to cause himself a lot of grief if he keeps insisting that a=v. a=dv/dt is certainly not a = v.

Zz.

yu_wing_sin
LeonhardEuler said:
Is it possible that the initial position is not zero? Could you explain to me exactly how you know that the result is wrong. I don't seem to see any contradiction within the data you gave.

Not right that. Many motions have the feature from still to accelerate, also the acceleration is average, for example the rockets or trains, in theoretical, it is possible. But often we will ignore the variable volecity for easy to calculate.

yu_wing_sin
ZapperZ said:
Yes, but he's going to cause himself a lot of grief if he keeps insisting that a=v. a=dv/dt is certainly not a = v.

Zz.

You mistook my meaning... Also this is the second paper of me in my scheme. Today now I found me is right, I can be brave to write it. Thank you.

Staff Emeritus
yu_wing_sin said:
You mistook my meaning... Also this is the second paper of me in my scheme. Today now I found me is right, I can be brave to write it. Thank you.

You are right about what, that the equation is "wrong"? Or did you not realize you were USING it wrongly?

I hate to think this is the "paper" you have been touting about in the other part of PF.

Zz.

Homework Helper
yu_wing_sin said:
Is this equation wrong?

This is a equation for
"uniform acceleration directed line motion at zero initial speed" in our textbook.

S=1/2 at^2

Here is the list to compare the answer:
t: 1 2 3
v: 2 4 6

If we use the equation to calculate, the answer is different to the list.

For example,

1/2(2)(3*3) = 9, s=9

but in the list, it shows the answer is 12!

Is my wrong? Or the equation is really wrong?

Even without the acceleration being given, you can look at the increase in 'v' each second and see that the velocity is increasing by 2 each second, which gives you an acceleration of 2 (which you correctly used in your equation).

Or you could differentiate the position equation and solve for acceleration (the acceleration is a little too obvious to bother with that, in this case).

Homework Helper
You repeatedly said "the list shows the answer is 12".

What list?? You didn't show us any list that include "12".

gunblaze
yu_wing_sin said:
Is this equation wrong?

This is a equation for
"uniform acceleration directed line motion at zero initial speed" in our textbook.

S=1/2 at^2

Here is the list to compare the answer:
t: 1 2 3
v: 2 4 6

If we use the equation to calculate, the answer is different to the list.

For example,

1/2(2)(3*3) = 9, s=9

but in the list, it shows the answer is 12!

Is my wrong? Or the equation is really wrong?

yu, the equation is not wrong.. However, it is your own working to find the distance to be 12, that is wrong.

Your problem lies where you took the velocity of each seconds and add them together... Argh..This is soo complicated to explain..,lets make life easier.Let me give you an example. Eg: You cannot assume that the velocity is at a constant at 4m/s when the time is at 2s. Let us take a look below.
0s------1s-----2s------3s------4s-------
Yes, u are right in saying that the velocity is 4m/s at the 2nd second but while it is moving towards the 3rd second, the velocity does not remain constant at 4m/s, it changes every single small portion of a second. Pls note that the acceleration is spreaded out throughout the line, but not just at every single seconds. Thus, 12m is the wrong displacement of this moving object here in this case.

Sry,.. not really good at explanation..

__________________

Physics was not meant to be static...

yu_wing_sin
Or I decided not to write this paper, because this paper has not any significant impact.