Acceleration in a parbolic path

A skier moves with a constant speed of 6 m/s along a parabolic path y = x²/20 . Find the acceleration when he is at (10,5)

2 Answers

we have

(dy/dt)²+(dx/dt)²=v²

but as

y=x²/20

dy/dt=(x/10)(dx/dt)

thus we get dx/dt interms of x ---- (i)

and differentiating again we get d²x/dt² in terms of x

further

dy/dt=(x/10)(dx/dt)

we know dx/dt from (i)

thus

d²y/dt²=1/10(dx/dt)² +x/10(d²x/dt²)

we get a_y and a_x

thus acceleration =

âˆš(a_x² + a_y²)

Ya...

V² = 36 = ( xV_x/10 )² + V_x² ----------------@

thn wat v get is :

A_y = V_x²/10 + xA_x/10 ----------------@

Applying x =10;
We can get

Ay = 1.8 + Ax; fir kya???

Now how to find A_y² + A_x²

Coz at the max we hav A_y in terms of A_x, A_x still remains a variable [7]

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