- #1

No1_129848

- 19

- 1

## Homework Statement

A cat rides a merry-go-round turning with uniform circular motion. At time t

_{1}= 2.00 s, the cat’s velocity is V

_{1}= (3.00 m/s)i + (4.00 m/s)j , measured on a horizontal xy coordinate system. At t

_{2}= 5.00 s, the cat’s velocity is V

_{2}= (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 t

_{2}- t

_{1}, which is less than one period?

## Homework Equations

[/B]

T = 2πr/V

a = V

^{2}/r

## The Attempt at a Solution

[/B]

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:

V

_{1}* V

_{2}= V

_{1x}* V

_{2x}+ V

_{1y}* V

_{2y}= 3*(-3) + 4*(-4) = -25

V

_{1}* V

_{2}= V

_{1}* V

_{2}CosΘ = √3

^{2}+4

^{2}*√(-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 = t

_{2}-t

_{1}= 5s-2s = 3s

T = 2Δt = 3s * 2 = 6s

T = 2πr/V → r = TV/2π = 6s * 5 m/s / 2π = 4,8 m

a = V

^{2}/r → (5 m/s)

^{2}/4,8 m = 5,2 m/s

^{2}

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:

a

_{avg}= V

_{2}- V

_{1}/ t

_{2}- t

_{1}

And I am not sure if I should consider just the magnitude of the two velocities, which means that a

_{avg}= 0, or if I should solve in unit vector notation, doing the following:

a

_{avg}= V

_{2}- V

_{1}/ t

_{2}- t

_{1}= (-3-3)i + (-4-4)j / 3s = -2i -2,67j → a

_{avg}= √(-2)

^{2}+(-2,67)

^{2}= 3,3 m/s

^{2}

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.

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