# Golf question

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1. Dec 18, 2014

### Arun Raja

1. The problem statement, all variables and given/known data
http://puu.sh/dzuq5/629a29dd48.png [Broken]

2. Relevant equations
x = vo t + ½ a t2
2 a x= v2 - vo2

3. The attempt at a solution
1/4 d = vo t + ½ a t2

multiply equation by 4,
d=4 ut+ 2a t2

so I am thinking answer is 4u, but it is 2u.

Last edited by a moderator: May 7, 2017
2. Dec 18, 2014

### 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 ?

3. Dec 18, 2014

### Arun Raja

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

4. Dec 18, 2014

### 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 = 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. Dec 18, 2014

### Staff: Mentor

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

6. Dec 18, 2014

### 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?