The solution for this problem is:
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters
v = dr/dt = [0.180t i - 0.0450t² j] m/s²
tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect
T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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I think it was either Robert Hook or Anton Van Leevwen Hoek
