Answer :
Answer:
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
Answer:
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.