Why does arc length tend to chord length in centripetal acceleration derivation?

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toforfiltum
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A comparison of corresponding parts of these two similar isosceles triangles yields

cc7ee24b-30e0-4cc9-80e2-47c7f4d1086b.gif

cce1bb03-f244-4e0f-a7d9-240c4cc0e010.gif

velocity triangle
displacement triangle

4562c164-1631-4934-9267-96a3a33872d1.gif


where

|-vo| = |vf| = v
In a small time interval Δt, the arc length s → c.


Okay, so I got this from Physics LAB, and I don't understand the last statement that states s tending to c. How could s even tend to c when the tangential velocity at A is not even directed towards c?
 
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toforfiltum said:
How could s even tend to c when the tangential velocity at A is not even directed towards c?

Do you understand what s→c means ?
 
Qwertywerty said:
Do you understand what s→c means ?
Not really, just that s is the arc and c is the cord. So I think that means the length of s gets very close to that of c.
 
toforfiltum said:
So I think that means the length of s gets very close to that of c.

Yes , it means that the length of the arc and the chord almost reach the same value ( obviously with arc greater than chord , but still getting close to the same value ) .
 
Qwertywerty said:
Yes , it means that the length of the arc and the chord almost reach the same value ( obviously with arc greater than chord , but still getting close to the same value ) .
Well, I don't get why the length of the arc tends to that of the chord.
 
toforfiltum said:
Well, I don't get why the length of the arc tends to that of the chord.

It doesn't , we just take such a value of arc and chord .
I'm sorry , do you know what a tangent is ?
 
Qwertywerty said:
It doesn't we just take such a value of arc and chord .
I'm sorry , do you know what a tangent is ?
A straight line that is perpendicular to one point in a circle.
 
Well, calculate the surface of triangle and compate it to the arc,you have an isocele triangle, the height of this triangle is h = r*cos(θ/2),It's surface is S = r2sin(θ)/2, now take this as θ→0 c = S/h = r2sin(θ)/(2*r*cos(θ/2)), use the fact that sin(θ) = 2*cos(θ/2)*sin(θ/2), c = r2sin(θ)/(2*r*(sin(θ)/2sin(θ/2)) = 2*r*sin(θ/2) = rθ = s, thus as θ → 0, c→s