Answer:
r=89.970 m
Explanation:
We need to calculate radius of curvature, using first and second derivatives of the functions x(t) and y(t).
From the given equation, we can get, that:
x(t)=1.89t^2 and y(t)=1.17t^3
Then, the first derivative:
x'(t)=2*1.89*t=3.78t and y'(t)=3*1.17*t^2=3.51t^2
The second derivatives are:
x''(t)=3.78 and y''(t)=3.51*2*t=7.02t
Equation for the radius of curvature, can be found as:
r=(x'^2+y'^2)^(3/2)/(x''y'-x'y'')
Note, that the denominator should be taken by its absolute value.
For the given time, we should calculate numerical values for the derivatives. For t=2.1 s:
x'(2.1)=7.938 m/s; x''(2.1)=3.78 m/s^2; y'(2.1)=15.4791 m/s and y''(2.1)=14.742 m/s^2
Using equation of the curvature radius and the values, we can get:
r=89.970 m
Answer:
ρ = 0.0657
Explanation:
We can apply the equation
ρ = 1/║r"(t)║
we have to get r"(t):
r'(t) = 3.78*t i + 3.51*t² j
r"(t) = 3.78 i + 7.02*t j
⇒ ║r"(t)║ = √(3.78²+(7.02*t)²)
If t = 2.1 s we have
║r"(2.1)║ = √(3.78²+(7.02*2.1)²) = 15.219
⇒ ρ = 1 / 15.219 = 0.0657 m
