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Is this equation wrong?

  • #1
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

  • #2
ZapperZ
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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.
 
  • #3
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?
 
  • #4
ZapperZ
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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.
 
Last edited:
  • #5
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.
 
  • #6
ZapperZ
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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

[tex] s = \frac{1}{2} at^2[/tex]?

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

[tex]s= s_0 + v_0t + \frac{1}{2} at^2[/tex]

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.
 
Last edited:
  • #7
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.
 
  • #8
ZapperZ
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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.
 
  • #9
ZapperZ said:
And you don't see ANYTHING wrong with plugging in the values of "v" into "a" in

[tex] s = 1/2 at^2[/tex]?

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

[tex]s= s_0 + v_0t + 1/2 at^2[/tex]

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".
 
  • #10
LeonhardEuler
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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?
 
  • #11
LeonhardEuler
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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.
 
  • #12
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If you're starting at postition s = 0 and time t = 0, the equation is just

[tex]s = v_{0}t + \frac{1}{2}at^2[/tex]

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

[tex]s = \frac{1}{2}at^2[/tex]

The [itex]s_{0}[/itex] and [itex]v_{0}[/itex] are just the displacement and velocity at time t = 0, respectively. What's hard to understand? ZZ's explanation seemed pretty clear to me :/.
 
  • #13
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
 
  • #14
ZapperZ
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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.
 
  • #15
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.
 
  • #16
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.
 
  • #17
LeonhardEuler
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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:
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v_1-v_0}{1-0}=2[/tex]
 
  • #18
LeonhardEuler
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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.
 
  • #19
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.
 
  • #20
ZapperZ
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LeonhardEuler said:
He seems to have gotten this from the fact that it is a uniform acceleration, so:
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v_1-v_0}{1-0}=2[/tex]
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.
 
  • #21
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.
 
  • #22
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.
 
  • #23
ZapperZ
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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.
 
  • #24
BobG
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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?
Your answer is correct!

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).
 
  • #25
HallsofIvy
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You repeatedly said "the list shows the answer is 12".

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

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