Bicyclist displacement, velocity and acceleration over 37 s

[intenzxboi]

Homework Statement

At one instant a bicyclist is 37 m due east of a park's flagpole, going due south with a speed of 14 m/s. Then, 37 s later, the cyclist is 37 m due north of the flagpole, going due east with a speed of 14 m/s. what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration? (Give all directions as positive angles relative to due east, where positive is measured going counterclockwise.)

The Attempt at a Solution

A) root (37^2 + 37^2) = 52.33

B) tan-1 (37/37)= 45 so 135 degrees

can someone tell me how to calculate the other questions??

 

[Redbelly98]

How are average velocity and acceleration defined?

 

[intenzxboi]

so basically for velocity is displacement/time
and acceleration is change in velocity/time

so for the magnitude of the velocity would it be 52.33/37
direction is the same??

how would i calculate the magnitude of the acceleration??

 

[intenzxboi]

k i manage to find velocity and acceleration

are the angles the same for the direction of velocity and acceleration?

 

[Redbelly98]

so basically for velocity is displacement/time
and acceleration is change in velocity/time

so for the magnitude of the velocity would it be 52.33/37
direction is the same??

Yes, that's the average velocity. It's in the same direction as the displacement, since we're just dividing displacement by 37 seconds to get the average velocity.

k i manage to find velocity and acceleration

are the angles the same for the direction of velocity and acceleration?

No. The average acceleration is calculated using the change in velocities, which go from due south (initially) to due east. That's different than the displacement, which started out due east and ended up being due north.

 

[intenzxboi]

since its going south and east at the same velocity wouldn't the angle be tan-1 (14/14)

so is the angle -135?

 

[Redbelly98]

I'm assuming an angle of 0 means due east here, and +90 degrees is due north ... please correct me if I'm wrong.

since its going south and east at the same velocity wouldn't the angle be tan-1 (14/14)

Yes.

so is the angle -135?

That's not the only angle whose tangent is 14/14 ...