# Kinematic in Quadratic Form Equation (missing something simple)

• Delta_Craig
In summary, the student threw her pencil upward at 18.3 m/s and it took it 2.86 seconds to first reach a point 12.2 m higher than where it was thrown. The student then incorrectly retained the square root sign and got two positive results from -18.3±29.36.
Delta_Craig
I just ran this pretty quick at work. But this is the general outline. Sorry for the slop, it will get better with time. Thanks in advance and any additional info can be supplied.

The material is from the Khan free course.

.....

A student is fed up with doing her kinematic formula homework, so she throws her pencil straight upward at 18.3 m/s.

How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?

t1= 0.869s
t2= 2.86s

My answers are negative time which is a neat concept. Doesn’t add up.

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There are a lot of numbers there. Maybe a bit of algebra first would help?

Delta_Craig
Agreed, amongst other mathematics.
But, it does seem pretty straight forward.
I’ll keep plugging at it, I know it’s an order of op that’s got me on the wrong path.

Delta_Craig said:
Agreed, amongst other mathematics.
But, it does seem pretty straight forward.
I’ll keep plugging at it, I know it’s an order of op that’s got me on the wrong path.

Which direction are you taking as positive? Up or down?

Delta_Craig
PeroK said:
Which direction are you taking as positive? Up or down?

Up for positive.
I have acceleration @ -9.81m/s

Delta_Craig said:
Up for positive.
I have acceleration @ -9.81m/s

Your first error is in line 2, where you have applied the same factor 1/2 to both g and s.
The next error is in multiplying those out you got a number near 1200 instead of near 120.
The result of that error was a negative number to square root, but instead of recognising something was wrong you just ignored the sign.
In your posted image, you then erroneously retained the square root sign having already substituted the root value.
Finally, you somehow got two positive results from -18.3±29.36.

haruspex said:
Your first error is in line 2, where you have applied the same factor 1/2 to both g and s.
The next error is in multiplying those out you got a number near 1200 instead of near 120.
The result of that error was a negative number to square root, but instead of recognising something was wrong you just ignored the sign.
In your posted image, you then erroneously retained the square root sign having already substituted the root value.
Finally, you somehow got two positive results from -18.3±29.36.

It's so easy to go wrong with numbers I always find. ##g## is one symbol; whereas ##9.81m/s^2## is eight symbols.

Why would anyone prefer eight symbols when one would do?

Delta_Craig said:
A student is fed up with doing her kinematic formula homework, so she throws her pencil straight upward at 18.3 m/s.

How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?

t1= 0.869s
t2= 2.86s

My answers are negative time which is a neat concept. Doesn’t add up.

Do they really teach you to do it that way? It seems extraordinary to me. I would never think of doing like that.

I just did, with up postive:

##s = v_0t - \frac 1 2 gt^2##

##\frac 1 2 gt^2 -v_0t + s = 0##

##t = \frac{v_0 \pm \sqrt{v_0^2 - 2gs}}{g}##

And I just plugged the numbers in a spreadsheet and out came the answer you quoted.

PeroK said:
Do they really teach you to do it that way? It seems extraordinary to me. I would never think of doing like that.

I just did, with up postive:

##s = v_0t - \frac 1 2 gt^2##

##\frac 1 2 gt^2 -v_0t + s = 0##

##t = \frac{v_0 \pm \sqrt{v_0^2 - 2gs}}{g}##

And I just plugged the numbers in a spreadsheet and out came the answer you quoted.
I’m sure you have a better grasp on this to make a statement like that, and I believe what your saying. I’m just going to run all the free courses (edit: knowing they are less than adequate, but a fair primer) I can before I start academically.

That’s the thing too, kinematic formulas are 100% new to me, so just learning these little gems is a lot of fun. I don’t have the perspective to know it’s not the most efficient way. I’m just hoping pencil on paper = progress

I’ll play with your formula tonight.

Delta_Craig said:
I’m sure you have a better grasp on this to make a statement like that, and I believe what your saying. I’m just going to run all the free courses (edit: knowing they are less than adequate, but a fair primer) I can before I start academically.

That’s the thing too, kinematic formulas are 100% new to me, so just learning these little gems is a lot of fun. I don’t have the perspective to know it’s not the most efficient way. I’m just hoping pencil on paper = progress

I’ll play with your formula tonight.

Or, try the calculator on your phone. Mine is a basic phone, but the calculator has an option to go into "scientific" mode. It's really easy to do formulas like that.

I’ll admit I rushed the formula at work, no excuse...

I’ll keep it cleaner from here on out, sorry for the waste of time on anyone’s part.

sat down and figured out the issue pretty quick.

On line 2, under the square root you are subtracting a velocity (m/s) from a squared velocity (m2/s2).

On line 3, the squared velocity is no longer squared. But you are about to take the square root of it.

On line 4, the square root of an un-squared velocity turns into an un-squared velocity.

If you are going to track units, track units.

Delta_Craig

## 1. What is a kinematic equation in quadratic form?

A kinematic equation in quadratic form is a mathematical representation of the relationship between position, velocity, acceleration, and time in a system with constant acceleration. It is derived from the basic kinematic equations, but is written in a quadratic form to solve for two unknown variables.

## 2. How do you solve a kinematic equation in quadratic form?

To solve a kinematic equation in quadratic form, you will need to have three of the four variables (position, velocity, acceleration, and time) known. Then, you can plug these values into the equation and use the quadratic formula to solve for the remaining unknown variable.

## 3. What is the significance of the coefficient of the squared term in a kinematic equation in quadratic form?

The coefficient of the squared term in a kinematic equation in quadratic form represents the acceleration of the system. This value is constant and can be used to determine the rate of change of velocity over time.

## 4. Can a kinematic equation in quadratic form be used for systems with non-constant acceleration?

No, a kinematic equation in quadratic form is only applicable to systems with constant acceleration. For systems with non-constant acceleration, a more complex equation, such as the Euler method, must be used to accurately represent the motion.

## 5. How is a kinematic equation in quadratic form related to projectile motion?

A kinematic equation in quadratic form can be used to solve for the vertical and horizontal components of a projectile's motion. By setting the vertical acceleration to -9.8 m/s^2 (the acceleration due to gravity), the equation can be used to analyze the projectile's vertical motion, while the horizontal component remains constant.

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