# I Kinematics Equations

1. Jun 22, 2017

### applepies

I would greatly appreciate it if someone can clarify the conditions on when to use to each of the kinematics equations, because I end up with the wrong answer for using the wrong equations of the 3.

1. vf = vi + at
When can I use this and when cannot?

2. vf^2 = vi^2 + 2a(xf-xi)
When can I use this and when cannot?

3. (xf-xi) = vi t + at^2/2
When can I use this and when cannot?

For example: Suppose a ball is thrown vertically upward from earth with velocity v and returns to its original height in time t. If the g value is reduced to g/6 then what would t be?

Why use equation 1 which will result in a time increasing by a factor of 6. If I use the second equation time will be increased by a factor of 6^1/2 ?

2. Jun 22, 2017

### Drakkith

Staff Emeritus
You can use any of these any time you have values for all of the variables except one.

We know that the ball will eventually come to a stop before reversing direction and accelerating back to Earth. We also know that the time it takes the ball to rise to its maximum height is the same as the time it takes it to fall back to Earth from that height. That means, depending on what values we know, we can solve for the time taken to either rise or fall and then double that value.

Looking at the equations and your problem, what variables do we have values for? What information do we know? We know the initial velocity, which immediately suggests that we should solve for the time taken for the ball to rise to its maximum height. Since we are solving for the rise, we know that the final velocity is zero. We also know the acceleration of gravity and the initial position, but we don't know the final position nor do we know the time. Looking at your equations, it looks like the only equation that we have a single unknown for is equation 1.

Both equation 2 and equation 3 have two unknowns each ($x_f$ and $t$), so we cannot use them.

3. Jun 22, 2017

### applepies

Thank you for the clarification!

Just to make sure I understand it this, if I was given the change in distance, the velocity, and acceleration, I would have to use (xf-xi) = vi t + at^2/2 to find t but I can't use vf = vi + at, because this doesn't account for the change in distance ?

4. Jun 22, 2017

### Drakkith

Staff Emeritus
That's right in general. But there are special cases. In your example above, we weren't given $v_f$ but we were able to use equation 1. That's because we were able to use zero for the final velocity because of the particular way the ball moves under gravity. In another situation you might not be able to do that and so you wouldn't be able to use equation one or two since you wouldn't have $v_f$ or $t$.

5. Jun 22, 2017

### applepies

Thank you so much! I feel like I have gotten through a small mound in the massive world of physics.

6. Jun 22, 2017

### Drakkith

Staff Emeritus
First step in solving any physics problem: determine what relevant information you're given.

7. Jun 22, 2017

### robphy

It should be clarified that these equations are valid only when you have constant acceleration.

The v's refer to "components of velocity" [or possibly vectors] but not speeds.
The a's refer to "components of accleration" [or possibly vectors] but not magnitudes-of-acceleration.

In addition, one should not mix-up x-components with y-components.
For example...
$\vec v_f = \vec v_i + \vec at$
or
$v_{x,f} = v_{x,i} + a_x t$
$v_{y,f} = v_{y,i} + a_y t$
Similarly,
$v_{x,f}^2 = v_{x,i}^2 + 2a_x(x_f-x_i)$
$v_{y,f}^2 = v_{y,i}^2 + 2a_y(y_f-y_i)$

Similarly,
$(x_f-x_i) = v_{x,i} t + a_xt^2/2$
$(y_f-y_i) = v_{y,i} t + a_yt^2/2$

..but never write, for example,
a horizontal displacement:
[generally false] $(x_f-x_i) = v_{x,i} t + (-g)t^2/2$
or
[generally false] $(x_f-x_i) = v t + (-g)t^2/2$, where $v=|\vec v|$.

8. Jun 30, 2017

### neilparker62

You could use something like this. Just write in the known variables and a ? for the unknown and line up against the ticks in the formula table to select the correct formula.

9. Jun 30, 2017

### Dr.D

Well, not really. These equations only apply when the acceleration is constant. That is not true in most real problems of interest.

10. Jun 30, 2017

### Drakkith

Staff Emeritus
Absolutely.