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

[Arun Raja]

Homework Statement

http://puu.sh/dzuq5/629a29dd48.png

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.

 

[BvU]

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 ?

 

[Arun Raja]

no. But I am not sure of the correct way too. so pls help .

 

[BvU]

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 = v_o t + ½ a t² . Correctly, if I assume v_o 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 ?

 

[NascentOxygen]

x = vo t + ½ a t2

Do you know a different equation relating distance, speed and acceleration?

 

[rude man]

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?