Solving Kinematics Questions with 5 Variables

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In summary: . 16.1 is wrong because when t=0 the vf will be zero so the d will be 0. for 16.7 the vf will be 16.7 when t=0 so the d will be 16.7
  • #1
MattJ81
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I've been relearning some stuff using khanacademy.org and I am completely lost on the kinematics exercises.

I have 5 variables, initial velocity final velocity acceleration time and distance. I am given 3 of them and have to solve for a 4th, which is fine until I have to deal with square roots that gives me 2 solutions.

I understand that I may have a negative time solution that isn't possible but some of the exercises produce 2 positive time solutions. Also I can have 2 velocity solutions and I can't figure out which one is right. If I plug my answers back in both answers seem fine.

Here is my latest problem...

kin3.jpg


As far as I can tell there is no solution because division by 0 is undefined. So what am I doing wrong?
 
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  • #2
MattJ81 said:
I've been relearning some stuff using khanacademy.org and I am completely lost on the kinematics exercises.

I have 5 variables, initial velocity final velocity acceleration time and distance. I am given 3 of them and have to solve for a 4th, which is fine until I have to deal with square roots that gives me 2 solutions.

I understand that I may have a negative time solution that isn't possible but some of the exercises produce 2 positive time solutions. Also I can have 2 velocity solutions and I can't figure out which one is right. If I plug my answers back in both answers seem fine.

Here is my latest problem...

kin3.jpg


As far as I can tell there is no solution because division by 0 is undefined. So what am I doing wrong?

Ah, if I read this correctly, a = 0, which seems silly and trivially easy but that's what you wrote. So what does your first equation look like?
 
  • #3
Yeah constant velocity, I don't have to fool with kinematics but how does he state that something divided by 0 can give an answer?

Heres the first ...

http://postimage.org/image/1xx6qw090/

I figured 16.1 or 16.7 but both answers appear to be plausible.

Here's the 2nd...

http://postimage.org/image/1yba67w4k/

I got this one wrong because I miscalculated the direction, is there an easy way to assume the direction like this? For example an initial velocity of 0 and a negative acceleration would yield a negative final velocity but with more complicated distances velocities and directions I have to think for a while about what's really going on.
 
  • #4
Welcome to PF!

Hi MattJ81! Welcome to PF! :smile:
MattJ81 said:
… some of the exercises produce 2 positive time solutions. Also I can have 2 velocity solutions and I can't figure out which one is right. If I plug my answers back in both answers seem fine.

There usually are two correct solutions, one is a "smash" and the other is a "lob", but they both take the same time to get to the same place. :wink:
As far as I can tell there is no solution because division by 0 is undefined. So what am I doing wrong?

You are using the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equation d = vit + 1/2 at2.

Here, a = 0, so the equation becomes simply d = vit,

to which the solution is t = d/vi. :wink:
 
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  • #5
your error is in trying to use the quadratic formula to solve an equation that is not quadratic!
 
  • #6


tiny-tim said:
There usually are two correct solutions, one is a "smash" and the other is a "lob", but they both take the same time to get to the same place. :wink:

How do I tell which one is going to be correct in these examples? I get em wrong every time

HallsofIvy said:
your error is in trying to use the quadratic formula to solve an equation that is not quadratic!

I see that, why in the world would the answer involve the quadratic formula??

The only problem I am having trouble with is deciding which of the 2 solutions is the one they are asking for, isn't there a way to intuitively check which one is right?

I appreciate the help
 
  • #7
If you look up the definition of the quadratic equation you will see that a,b and c must be constants and "a" must not be equal to 0. You cannot use the quadratic equation in your problem when the acceleration is 0. As been pointed out your equation is linear.
 
  • #8
Makes complete sense, Ill ignore this as a problem with the site itself.

Take a look at the problem with solutions 16.1 and 16.7. What makes 16.1 wrong?
 
  • #9
d=Vi*t-.5*a*t^2 NOT Vf*t-.5*a*t^2
 
  • #10
Hi MATTJ81

the problem in your first question is that when the acc is 0 the equation itself trasforms to s=ut
 

Related to Solving Kinematics Questions with 5 Variables

1. How do I determine the five variables in a kinematics problem?

The five variables in a kinematics problem are displacement, velocity, acceleration, time, and initial velocity. Displacement is the change in position, velocity is the rate of change of displacement, acceleration is the rate of change of velocity, time is the duration of the motion, and initial velocity is the velocity of the object at the beginning of the motion.

2. What equations can I use to solve kinematics problems with five variables?

The most commonly used equations for solving kinematics problems with five variables are the equations of motion:
- s = ut + ½at2 (for displacement)
- v = u + at (for final velocity)
- a = (v - u)/t (for acceleration)
- t = (v - u)/a (for time)
- s = (u + v)/2 * t (for average velocity)
where s is displacement, u is initial velocity, v is final velocity, a is acceleration, and t is time.

3. Can I use these equations for any type of motion?

Yes, these equations can be used for any type of motion as long as the acceleration is constant. This means that the velocity is changing at a constant rate over time. If the acceleration is not constant, more advanced equations or techniques may be required to solve the problem.

4. How do I determine which equation to use for a specific kinematics problem?

First, identify which variables are given in the problem and which one is being asked for. Then, look at the equations and see which ones contain the given and asked for variables. Choose the equation that contains all the given variables and the asked for variable. If there are multiple equations that meet this criteria, use the one that is easiest to solve or the one that gives the most accurate answer.

5. Can I solve kinematics problems with more or less than five variables?

Yes, the equations of motion can be used for problems with more or less than five variables. However, if there are more variables, additional equations or techniques may be needed to solve the problem. If there are less than five variables, it may not be possible to find a unique solution, or the solution may not accurately represent the motion.

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