#1

A hawk flies in a horizontal arc of radius 13.0 m at a constant speed of 4.25 m/s.

(a) Find its centripetal acceleration.

I got 1.3894 m/s^2 which is right.

(b) It continues to fly along the same horizontal arc but increases its speed at the rate of 1.20 m/s2. Find the acceleration (magnitude and direction) under these conditions.

__________m/s^2 @ _______ degrees (towards the center of the circle)

I'm confused about part b...I've read about radial and tangential accelerations but i'm still very confused

we're doing harmonic motion and oscillations so i kinda forgot

Thanks! anything would help

(a) Find its centripetal acceleration.

I got 1.3894 m/s^2 which is right.

(b) It continues to fly along the same horizontal arc but increases its speed at the rate of 1.20 m/s2. Find the acceleration (magnitude and direction) under these conditions.

__________m/s^2 @ _______ degrees (towards the center of the circle)

I'm confused about part b...I've read about radial and tangential accelerations but i'm still very confused

we're doing harmonic motion and oscillations so i kinda forgot

Thanks! anything would help

#2

wat?

#3

well acceleration is a vector. so just separate out all of your X and Y coordinations and calculate those individual accelerations then just add the vectors together.

#4

i'm doing the exact same crap in physics... i'm lost... need loads of help myself.

#5

lol wut?

fixed

#6

It's 42. Everyone knows hawks fly at 42.

#7

well, centripetal acceleration (alpha) is what you're trying to find I guess, and you're given linear acceleration (a).

Use the formula:

a = (alpha)(radius)

Use the formula:

a = (alpha)(radius)

#8

Actually, I think you may have to use:

(total acc)^2 = (linear acc)^2 + (cent. acc)^2

(total acc)^2 = (linear acc)^2 + (cent. acc)^2

#9

lol deffinetly mastering physics isnt it...

damn i gotta do mine soon.. stupid electric fields==integrals

stupid engineering..

[edit] if tis a constant radius the tangential acc should be 90 degrees and the radial should be 0 degrees should it not? and hte total accelleration would be cos(radial accel) + sin (tangential accel) am i not correct?..

damn i gotta do mine soon.. stupid electric fields==integrals

stupid engineering..

[edit] if tis a constant radius the tangential acc should be 90 degrees and the radial should be 0 degrees should it not? and hte total accelleration would be cos(radial accel) + sin (tangential accel) am i not correct?..

*Last edited by JimmyBanks6 at Feb 22, 2009,*

#10

Actually, I think you may have to use:

(total acc)^2 = (linear acc)^2 + (cent. acc)^2

hey that's right.

i don't know why i've never seen that before

my only problem how do you get the angle? is that also with vectors?

#11

edit my last edit

i think this edit is correct --> the angle should be theta=tan^-1[(tang accel) / (rad accel)]

i may be incorrect as i am not writing this down and trying to think at through in my head

i think this edit is correct --> the angle should be theta=tan^-1[(tang accel) / (rad accel)]

i may be incorrect as i am not writing this down and trying to think at through in my head

*Last edited by JimmyBanks6 at Feb 22, 2009,*

#12

it's actually (rad accel/ tan accel) apparently.

#13

it's actually (rad accel/ tan accel) apparently.

hmm that does make sense.. i probly shoulda written it down and i wulda made my way to that conclusion.. sry if i messed up ur Mastering Physics... which i think it is your working on?

#14

Just tell your teacher this-

Your mother is so fat, when she stepped on a scale her acceleration from rest resulted in a force between the two conductors of 6.02 circum-plex 23 newton's priminant and magnetic flex around the closed curve was proportional to the algebraic sum of electric currents flowing through that closed curve.

But, on topic. Idklol. Sorry I can't help.

Your mother is so fat, when she stepped on a scale her acceleration from rest resulted in a force between the two conductors of 6.02 circum-plex 23 newton's priminant and magnetic flex around the closed curve was proportional to the algebraic sum of electric currents flowing through that closed curve.

But, on topic. Idklol. Sorry I can't help.