# Bike riding using an angle

1. Sep 25, 2011

### physicsgurl12

1. The problem statement, all variables and given/known data
riding his bike with v=(8.4m/s,25 degrees north of east) for ten minutes. how far to the north of the starting position did he end up?

2. Relevant equations

x=v^2sin2angle/g

3. The attempt at a solution
70.56sin50/g
theres not even a t in this.

2. Sep 25, 2011

### cepheid

Staff Emeritus
This equation is for projectiles.

The guy on his bike is moving at a constant speed. That makes things really easy. You have speed and time, how do you find distance? For sure, you know this.

Assuming you find this distance, it is the distance travelled in the direction of motion (which is 25 degrees north of east in this case). How, then, do you break that down into total northward displacement and total eastward displacement? Now it's just geometry. Draw a diagram of the situation, this will help immensely.

3. Sep 25, 2011

### physicsgurl12

v=d/t 8.4m/s=d/600s d=5040m ? draw a traingle??

4. Sep 25, 2011

### cepheid

Staff Emeritus
Yes, exactly. 5040 m is the distance travelled along the diagonal line that is oriented 25 degrees above the horizontal. (On a diagram where N-S is drawn to be vertical and E-W is drawn to be horizontal).

5. Sep 25, 2011

### physicsgurl12

so about 2100 d?

6. Sep 25, 2011

### cepheid

Staff Emeritus
Post what you did to arrive at that answer and why you think that is the thing to do...(remember it's whether or not you understand the concept and the methods of your solution that are important, not the final numerical answer).

7. Sep 25, 2011

### physicsgurl12

okay well i was kinda confused on what goes where on the triangle

8. Sep 25, 2011

### cepheid

Staff Emeritus
Well that still doesn't tell me what you did.

Anyway, the correct answer is just over 2100 m, so you probably did the right thing. I just wanted to make sure you understood the geometry and why that was the method of solution before confirming your answer.

9. Sep 25, 2011

okay got ya.