Question

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.

a(t) = e^(2t) i + sqrt(t) j + 2t k
with v(0) = 3i + 0.5j + 2k and r(0) = i.

Answer 1

The velocity vector of the particle is (e^(2t) - 1) i + (2/3) t^(3/2) j + t^2 k, and its initial velocity is 0.

Step-by-step explanation:

The velocity vector of a particle can be found by integrating the acceleration vector with respect to time. In this case, the acceleration vector is given by a(t) = e^(2t) i + sqrt(t) j + 2t k.

Integrating the acceleration vector, we get the velocity vector: v(t) = (e^(2t) - 1) i + (2/3) t^(3/2) j + t^2 k.

Given that v(0) = 3i + 0.5j + 2k, we can substitute t = 0 into the velocity vector to find the initial velocity of the particle: v(0) = (e^0 - 1) i + (2/3) * 0^(3/2) j + 0^2 k = 0i + 0j + 0k = 0.

Therefore, the velocity vector of the particle is v(t) = (e^(2t) - 1) i + (2/3) t^(3/2) j + t^2 k, and its initial velocity is 0.