Solving an Equation: x = vo t + ½ a t2

  • Thread starter Arun Raja
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    Kinematics
  • #1
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Homework Statement


http://puu.sh/dzuq5/629a29dd48.png [Broken]

Homework Equations


x = vo t + ½ a t2
2 a x= v2 - vo2[/B]


The Attempt at a Solution


1/4 d = vo t + ½ a t2[/B]
multiply equation by 4,
d=4 ut+ 2a t2

so I am thinking answer is 4u, but it is 2u.
 
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  • #2
Does that mean you think ½ a t2 in the ¼ d case is the same as 2 a t2 (different t ?) in the distance =d case ?
 
  • #3
no. But I am not sure of the correct way too. so pls help .
 
  • #4
Well, if we go through "all variables and given/known data" in "problem statement" and "relevant equations", we see u, d, x, v0, a, t and v.

In your attempt at solution, you have ¼ d = vo t + ½ a t2 . Correctly, if I assume vo is u .

You don't use the second relevant equation. Why not? What does it mean ? What is v ?

All this is for the ¼ d case. What changes for the d case ? And what stays the same ?
 
  • #5
x = vo t + ½ a t2
Do you know a different equation relating distance, speed and acceleration?
 
  • #6
Look at this in energy terms instead. It took kinetic energy E to get to d/4, so how much energy is needed to get to d?
And finally, what is the relation between kinetic energy and speed?
 

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