Quadratic Equations and kinetics

In summary, To find the time required for a train to reach 100m, given an initial velocity of 20m/s and an acceleration of 0.5m/s2, use the kinematic equation \Delta x = v_0t + \frac{1}{2}at^2 and rearrange it to get the quadratic equation 0.25t^2 + vt - \Delta x = 0. From there, you can either manually factorize it or use the quadratic formula to solve for time.
  • #1
mode656
14
0

Homework Statement


Find the time required for a train to reach 100m if V0=20m/s and acceleration is
0.5m/s2


Homework Equations


s=v0t+(1/2)at2


The Attempt at a Solution


i am not sure of how to go about getting s=(20)t+(1/2)0.5t2 into quadratic form for use in the quadratic formula
this is my solution without the quad formula
s=v0t+(1/2)at2
100=(20)t+(1/2)0.5t2
5=t+(1/2)0.5t2
5=t+.25t2
5=t3
cube root of 5 is 1.71 which is def wrong:(
 
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  • #2
mode656 said:
5=t+.25t2
5=t3
cube root of 5 is 1.71 which is def wrong:(

Just bring the '5' on the other side to get

0.25t2+t-5 = 0

at2+bt+c=0
 
  • #3
in my book its sets up the quadratic form as t2+80t-400=0
the answer is still wrong i get 10??

it is supposed to be 4.5
 
  • #4
[tex]\Delta[/tex]x=vit+[tex]\frac{1}{2}[/tex]at2

There's your kinematic equation.

Edit: sorry didn't see your attempt.

[tex]\Delta[/tex]x=vit+[tex]\frac{1}{2}[/tex]at2

0=vit+[tex]\frac{1}{2}[/tex]at2-[tex]\Delta[/tex]x

[tex]\frac{1}{2}[/tex]at2+vit-[tex]\Delta[/tex]x=0

Quadratic form
 
  • #5
i know that's the equation but i need the quadratic formula to solve for time when only acceleration, distance and initial velocity is known
 
  • #6
mode656 said:
i know that's the equation but i need the quadratic formula to solve for time when only acceleration, distance and initial velocity is known

Those are the only variables you need to know. I edited my above post, sorry for reposting what you knew.
 
  • #7
don't worry about it. My problem does not lie within my use of the quadratic formula. My question is just pretty much how the book got t^2+80t-400=0
from 100=(20)t+(1/2)0.5t2
 
  • #8
mode656 said:
don't worry about it. My problem does not lie within my use of the quadratic formula. My question is just pretty much how the book got t^2+80t-400=0
from 100=(20)t+(1/2)0.5t2

Multiply your equation throughout by 4.
 
  • #9
any specific reason why? or is it just so?
 
  • #10
mode656 said:
any specific reason why? or is it just so?

If you plan to manually factorize it, it is easier to find the factors.

So instead of factorizing (1/4)t2+20t-100=0, it is easier to factorize out t2+80t-400.

If you plan to use the quadratic equation formula, then it really does not matter what form you put it in.
 

1. What are quadratic equations and how are they solved?

Quadratic equations are equations that contain a variable to the second power, such as x^2. They can be solved by using the quadratic formula, factoring, or completing the square.

2. How are quadratic equations used in kinetics?

Quadratic equations are used in kinetics to model the motion of objects. The position, velocity, and acceleration of an object can be described by quadratic equations.

3. What is the difference between a linear and quadratic equation?

A linear equation has a variable raised to the first power, while a quadratic equation has a variable raised to the second power. This means that a quadratic equation will have a parabolic curve, while a linear equation will have a straight line.

4. Why are quadratic equations important in chemistry?

In chemistry, quadratic equations are used to describe the rate of a chemical reaction. The rate law for many reactions involves a quadratic relationship between the concentration of reactants and the rate of the reaction.

5. How do you graph a quadratic equation?

To graph a quadratic equation, you can plot points by substituting different values for the variable and then connecting the points with a smooth curve. Alternatively, you can use the vertex and a few other key points to create a parabola.

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