# Homework Help: Kinematics Question

1. Sep 3, 2011

### MattJ81

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...

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

2. Sep 3, 2011

### JeffKoch

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. Sep 3, 2011

### MattJ81

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 whats really going on.

4. Sep 3, 2011

### tiny-tim

Welcome to PF!

Hi MattJ81! Welcome to PF!
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.
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.

Last edited by a moderator: Apr 26, 2017
5. Sep 3, 2011

### HallsofIvy

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

6. Sep 4, 2011

### MattJ81

Re: Welcome to PF!

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

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. Sep 4, 2011

### RTW69

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. Sep 4, 2011

### MattJ81

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. Sep 4, 2011

### RTW69

d=Vi*t-.5*a*t^2 NOT Vf*t-.5*a*t^2

10. Sep 4, 2011

### Shivam123

Hi MATTJ81

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