A cat rides a merry-go-round turning with uniform circular motion. At time t1 = 2.00 s, the cat’s velocity is V1 = (3.00 m/s)i + (4.00 m/s)j , measured on a horizontal xy coordinate system. At t2 = 5.00 s, the cat’s velocity is V2 = (3.00 m/s)i + (4.00 m/s)j.
What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s average acceleration during the time interval t2 - t1, which is less than one period?
T = 2πr/V
a = V2/r
The Attempt at a Solution
So, the first thing I did was sketch the situation, and the vectors seem to be on the opposite end of the circumference, so i tested for that:
V1 * V2 = V1x * V2x + V1y * V2y = 3*(-3) + 4*(-4) = -25
V1 * V2 = V1 * V2 CosΘ = √32+42*√(-3)2+(-4)2 = 25 CosΘ → CosΘ = -1 → Θ = Cos-1(-1) = 180°
So it's proven that the cat is in two different positions that cover half of the circumference, which means that
Δt = ½T
Δt = t2-t1 = 5s-2s = 3s
T = 2Δt = 3s * 2 = 6s
T = 2πr/V → r = TV/2π = 6s * 5 m/s / 2π = 4,8 m
a = V2/r → (5 m/s)2/4,8 m = 5,2 m/s2
So this should answer the first part of the problem, now I'm asked to find the average acceleration, and I'm not sure what approach is the correct one, I know the equation for the average acceleration is:
aavg = V2 - V1 / t2 - t1
And I am not sure if I should consider just the magnitude of the two velocities, which means that aavg = 0, or if I should solve in unit vector notation, doing the following:
aavg = V2 - V1 / t2 - t1 = (-3-3)i + (-4-4)j / 3s = -2i -2,67j → aavg = √(-2)2+(-2,67)2 = 3,3 m/s2
What is the correct way to answer the second question?
(also, kinda off topic, is there a way to write fractions and vectors on the forum?)
Thanks in advance, Ivan.