The box of negligible size is sliding down along a curved path defined by the parabola y=0.4x2. When it is a A(xA = 2 m, yA = 1.6 m), the speed is v = 8 m/s and the increase in speed is dv/dt = 4 m/s2

a=8.06m/s^2

Explanation:

The box can be considered negligible body slidding down along  a curved path defined by the parabola Y=Ax^2

Note:

When it's at A(x=2m, y=1.6m),

the speed Vb=8m/s and the increase in speed=4m/s^2

To find the acceleration,

Y=Ax^2

dy/dx=8x

d^2y/dx^2=8

p={[1+(dy/dx)^2]^3/2}/|d^2y/dx^2| 1

substituting into 1, we have

p=8.39624m

an=v^2/p

an=8^2/8.39624=7.6224m/s^2

a=sqrt(at^2+an^2)

a=sqrt(4^2+7.62246^2)

a=8.06m/s^2

acceleration = 8²/8.4 = 7.62m/s².

Explanation:

Given that at A

x = 2m, y = 1.6m

y = 0.4x²

dy/dx = 0.8x

At x = 2m

This acceleration is that of a body around a circular path of radius of curvature r. Since the position is given by a function y(x). We will use calculus to get the radius and then find the acceleration a = v²/r

The radius of curvature of the path from calculus is given by

r=([1+(dy/dx)²]^(3/2))/(d²y/dx²)

dy/dx = 0.8×2 = 1.6

d²y/dx² = 0.8

r = ([1+(1.6)²]^(3/2))/(0.8) = ([1+ 2.56]^1.5)/0.8 = 3.56^1.5/0.8 = 6.72/0.8 = 8.4m

r = 8.4m

acceleration = v²/r

At this instant v = 8m/s

acceleration = 8²/8.4 = 7.62m/s².

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. west