3) The position vector of a particle moving in three dimensions is given by r= 3

Question

3) The position vector of a particle moving in three dimensions is given by r= 3 t 1-227 + 2 k, where t is the time in seconds and all numerical coefficients have the proper units so that r is in meters. (a) Find the magnitude of the particle's position vector as a function of time, and then find its value when t = 2 s. (b) Find the particle's velocity vector u, as a function of time, and find its magnitude and direction at t = 2 s. (c) Find the particle's acceleration vector a as a function of time, and find its magnitude and direction at 2 s.

Explanation / Answer

a) magnitude of r = sqrt((3t)^2 + (-2t^2)^2 + 2^2)

= sqrt(9t^2 + 4t^4 + 4)

at t =2

magnitude = sqrt(9*2^2 + 4*2^4 + 4)

= 10.2 m

b) v = dr/dt

= 3i - 4tj

v at t = 2

= 3i - 4*2 j = 3i -8j

magnitude = sqrt(3^2 + 8^2) = 8.54 m

c) a = dv/dt

= -4j

magnitude at t = 2  

4 m/s2

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