Asked by sarah
the velocity of a particle is given by v=[16t^2i+4t^3j +(5t+2)k]m/s, where t is in seconds. If the particle is at the origin when t=0, determine the magnitude of the particle's acceleration when t=2s. What is the x,y,z coordinate position of the particle at this instant.
Answers
Answered by
drwls
The acceleration is
dv/dt = 32t i + 12 t^2 j + 5 k
Plug in t = 2 to get the acceleration vector at t =2.
The position vector is the integral of v(t),
r(t) = [(16/3)t^3 + a]i +
(t^4 + b)j + [(5/2)t^2 + 2t + c] k
where a, b and c are constants of integration. Use the position at t = 0 to solve for them. It appears that a=b=c=0
dv/dt = 32t i + 12 t^2 j + 5 k
Plug in t = 2 to get the acceleration vector at t =2.
The position vector is the integral of v(t),
r(t) = [(16/3)t^3 + a]i +
(t^4 + b)j + [(5/2)t^2 + 2t + c] k
where a, b and c are constants of integration. Use the position at t = 0 to solve for them. It appears that a=b=c=0
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