What equation to use for simple kinetics quesitons?

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For solving simple kinematics questions, the discussion emphasizes using standard constant acceleration equations. Participants suggest starting with the known values and matching them to the appropriate equations, such as \( v = u + at \) and \( s = ut + \frac{1}{2}at^2 \). There is a focus on deriving the equations correctly, particularly when solving for time using the quadratic formula. The importance of substituting back into the original equations to verify correctness is also highlighted. Understanding how to identify and apply the right equations is crucial for solving kinematics problems effectively.
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For the questions below, which kinematic equations would I use? I have already tried these on my own but can not find a way to match the answers with the answer sheet, the answers on the answer key are RIGHT, as they are approved by my teacher.
Okey so the questions are
1) A baseball is thrown up vertically at 25m/s. What velocity will the ball have after:
a)1.0s
b)2.0s
c)3.0s

2) A soccer ball is kicked straight up at a speed of 30m/s. How long will it take to reach a height of 5.0m?
It would be great if you showed me how you got your answers, if not, I will really appreciate the equations, THANKS!
 
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Welcome to PF!

Hi Alipepsi! Welcome to PF! :wink:

For both questions, use one of the standard constant acceleration equations.

If you're still stuck, show us what you've tried, and then we'll know how to help! :smile:
 
Thanks
 
Tiny, I'am confused =(
I've looked over the equations but can not figure out what equation to use. PLEASE HELP ME I have a test after march break =S
 
For 1), you have u t and a, and you want v

For 2), you have u s and a, and you want t

So which of the constant acceleration equations has only u t a and v,

and which has only u t a and s ? :smile:
 
a=vf-vi/t

Then just derive it :P
 
Yup! :biggrin:

(And the other one? :smile:)​
 
Vf^2=vi^2+2a(d)
v=sq root 30^2-2(9.81)(5)
v= 28.3m

THEN

Solve for time using the equation: t=(v-vf)/a

answer is 0.17seconds ;D

THANKS FOR YA HELP MAN! :D
 
You're welcome! :smile:

But quicker would have been s = ut + 1/2 at2 :wink:
 
  • #10
I see, can you help me with how to know what equations to use for questions. Like I was confused for what equation to use for the second problem. Do I look for all the values I'am given with the values in all teh equatiosn then pick the one? Ik my quesitons is silly :P
 
  • #11
I always write down a list of all the given values when I start a problem. Then I compare them to the equations that I know.
 
  • #12
tiny-tim said:
You're welcome! :smile:

But quicker would have been s = ut + 1/2 at2 :wink:

So you have to derive for t right?

This is what I got:

t= sq root s-ut/0.5a

This is correct? Really need to know this asap
 
  • #13
yes, you'd have to solve for t
 
  • #14
SHISHKABOB said:
yes, you'd have to solve for t

did I derive for t correctly there?
 
  • #15
you'd have to use the quadratic equation to solve for t in that case.

notice how t= sq root s-ut/0.5a still has a t ion both sides of the equation.
 
  • #16
SHISHKABOB said:
you'd have to use the quadratic equation to solve for t in that case.

notice how t= sq root s-ut/0.5a still has a t ion both sides of the equation.

If I was to use quadratic, then how would I get my A, B, and C values from solving that equation?
 
  • #17
well the equation is also this:

1/2 at2 + ut - s = 0

so if t is the variable, what are the constants?
 
  • #18
Alipepsi said:
did I derive for t correctly there?

You can check for yourself by substituting your formula for t back into the original equation to see if everything cancels out algebraically. Or you can use it to calculate t, then substitute it and your given quantities back into the original equation to see if everything cancels out numerically.
 
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