Calculating Average Velocity and Acceleration of the Singapore Flyer

[gjh]

Started over. Back to vectors. v =.26[cos 60i+sin 60j) = .13i+.22j (this is instantaneous velocity). avg a = delta v/delta t = (.13i+.22j-.26i)/300 = (-.13i+.22j)/300 = (-4.3i+8.7j)10^-4 - this is closer, but magnitude of a would be 9.7(10)^-4 ... which is way off from 9(10)^-4. Almost there ... kinda/sorta. Comments?

 

[haruspex]

Started over. Back to vectors. v =.26[cos 60i+sin 60j) = .13i+.22j (this is instantaneous velocity). avg a = delta v/delta t = (.13i+.22j-.26i)/300 = (-.13i+.22j)/300 = (-4.3i+8.7j)10^-4 - this is closer, but magnitude of a would be 9.7(10)^-4 ... which is way off from 9(10)^-4. Almost there ... kinda/sorta. Comments?

Looks like rounding error.
Don't be so quick to plug in numbers. Keep everything algebraic as long as you can, only plugging in numbers at the end. You should find it simplifies a lot and gives an exact result,

 

[gjh]

Yes, plus I made a stupid computational error. .22j/300 = 7.33*(10)^4 not 8.7jx(10)-4. Thus, my answer would be a (average) =(-4.3i+7.3j)10^-4 in component form. mag a then is sqrt [(-4.3)^2+(7o.3)^2(10^-4)] = 8.9*10^-4 m/s^2. This is very close to book answer of -4.4i+7.6j)*10^-4 with mag 8.8 m/s^2. So rounding error was part of the problem - and carelessness. Thx for the tip!