# Windsurfer making a turn (Centripetal Acceleration)

1. Apr 2, 2009

### eddyy08

"A windsurfer and their board have a total mass of 100 kg. While making a turn downwind, their heading changes from due west to due south over a 5 s time interval. If they maintain a constant speed of 10 m/s while doing this, determine their acceleration while taking the turn."

v=Δx/Δt
10=(2πr/4)/5 assuming his path is a perfect 1/4 of a circle

r=31.8 m

a=v^2/r
a=10^2/31.8

It seems like I've done it correctly, but 3.14 or π doesn't seem correct to me.... please verify if my answer is correct.

2. Apr 2, 2009

### LowlyPion

I think you've got things a little mixed. So let's start with this. In terms of linear acceleration it's 0. Surferdude maintained constant speed.

But in terms of the velocity vector v there was an acceleration due to his change in direction. So for acceleration a, we have ...

a = Δv/Δt = Δ<Vx,Vy>/Δt.

Over the 5 sec time period you have

a = Δv/Δt = Δ<10 x,10 y>/5 = Δ<2 x, 2 y>. Directed at 45° to positive x. (East)

3. Apr 2, 2009

### eddyy08

I know that the linear acceleration is 0, what I want to figure out is the centripetal acceleration.

4. Apr 2, 2009

### LowlyPion

In thinking about it, I guess I'm the one that had it mixed. It does seem to indicate that they want the acceleration "while" making the turn not "in" making the turn, and centripetal acceleration is = π as you already found, directed radially.

As the numbers work out you end with R = V2/π which means that V2/R = π

5. Apr 2, 2009

### eddyy08

Thanks a lot I appreciate it :)