a = v^2/r
a = sqrt(10^2 + 6^2) = sqrt(136)
Distance particle has moved = (angle moved through)*radius =
(pi/2)*r
Distance = velocity*time = v*t
v*t = (pi/2)*r
Using t = 2 we have v = (pi/4)*r
a = [(pi/4)*r]^2
r = a/[pi^2/16] = 16*a/pi^2 = 16*sqrt(136)/pi^2
r = 18.90 meters
Chapter 04, Problem 063 At t, -5.00 s, the acceleration of a particle moving at constant...
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At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vector v i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 3.70 s, the particle's velocity is vector v = (7.40 i + 6.90 j) m/s. (Use the following as necessary: t. Round your coefficients to two decimal places.) (a) Find the acceleration of the particle at any time t. vector a = m/s2 (b)...
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The figure gives the acceleration a versus time t for a particle
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Chapter 02, Problem 018 (e) the acceleration of the particle at 4.00 s. (d) What is the maximum positive coordinate reatn of the particdle at the instant the particle is not moving maximum positive velocity reached by the particle and (g) (other than at t 0)? () Determine the average velocity of the particle between t 0 and t -4.00s. s is given by x-13.02-4.00r, where x is in meters and t is in seconds. Determine (a) the position, (b)...