How Far North Did the Biker Travel in 10 Minutes with Given Velocity and Angle?

[physicsgurl12]

Homework Statement

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?

Homework Equations

x=v^2sin2angle/g

The Attempt at a Solution

70.56sin50/g
there's not even a t in this.

 

[cepheid]

x=v^2sin2angle/g

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 traveled 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.

 

[physicsgurl12]

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

 

[cepheid]

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

Yes, exactly. 5040 m is the distance traveled 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).

 

[physicsgurl12]

so about 2100 d?

 

[cepheid]

so about 2100 d?

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).

 

[physicsgurl12]

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

 

[cepheid]

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

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.

 

[physicsgurl12]

okay got ya.