# Equations of Motion question

1. Oct 23, 2011

### jaxreid

1. The problem statement, all variables and given/known data
A speed skater crosses the start line of a straight 200m downhiull course with speed of 30m per second. She accellerated uniformly all the way down taking 5 sec to cover the course - what is her speed at the finish line? I know this should be simple, but just can't get the answer to work out the same as the solution in the paper.

2. Relevant equations

s = 1/2 (u+v)t

3. The attempt at a solution
200 = 1/2 (30+v)5
200/5 = 1/2(30+v)
40 = 1/2(30+v)
2 x 40 = 30 + v
80 - 30 = v
v = 50m per sec

Answer in book is 45 - HOW??????????????????

Last edited: Oct 23, 2011
2. Oct 23, 2011

### LawrenceC

I have to wonder about the answer you are given. According to them the average speed is .5(30+45)=37.5 m/s. That won't cover the 200 m in 5 seconds. I agree with you.

The constant (uniform) acceleration rate is 4 m/s*s gotten from: x = V0*t+.5*a*t^2

Inserting that into Vf = V0 + a*t yields Vf=50 m/s.

So I don't understand either.

3. Oct 23, 2011

### jaxreid

Lawrence - many thanks - was starting to think I'd lost the plot, am glad you agree. The question's from a past paper and the published answer was 45. Just goes to show, they don't always get it right, will point this out to my physics teacher when school returns on Tuesday. Thanks again for your prompt reply.

Jax