Solving Quadratic Equation: Find the Time

Click For Summary
The discussion revolves around solving a quadratic equation related to time in motion, specifically using the formula x - xo = Vo t + 1/2 at^2. The user initially finds two positive time solutions, t=40.28 and t=40.58, and questions which one to choose when considering negative acceleration. It is clarified that both times can be valid in scenarios like projectile motion, where an object passes the same position twice—once on the way up and once on the way down. The conclusion drawn is that the lower time value typically represents the first instance the object reaches that position, while the second time indicates when it returns after deceleration. Understanding the context of motion is crucial in determining the correct time solution.
intenzxboi
Messages
98
Reaction score
0
k so i used
x-xo= Vo t + 1/2 at^2

plugged in all my values
then moved everything to one side and used the quadratic formula

here is my question.

when i solve for it i get
t=40.28 and t= 40.58

So which t is my time?
 
Physics news on Phys.org
Er.. well, you seem to not have included the actual question?
 
lets just use imaginary numbers

if u solve for t and get one + and one - answer u know time can't be negative if you are traveling forward.. so the answer is the + one

but what if your t comes out to be both + and you are going foward with -acceleration
 
This isn't something that has a general rule - sometimes both of them can be the correct time and other times there's some condition that rules the other one out.

Looking at your equation now it's the one used for 2D motion.. Taking that as an example, if you're considering a projectile and get two times it should be because the object is actually at the same horizontal position twice - once when going up and once when going down. It depends on what you want to calculate, in that case.

Probably this is no help, but I honestly can't do much more. Sorry ;/
 
I actually got it... had to think about it for a while..
my guess is that its the lower time because that's when the object gets there first...
the second time is when v=0 and the -acc. is making it go backwards so it goes back past the same position

right?
 
Yeah, if you're considering something that's thrown straight up into the air that would be the case, I'd say.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K