Find the position vector of a particle given a(t)= i + 2 j + 2t k, the initial velocity is zero, and the initial position at t=0 is (1,0,1)

Answer 1

We can start by integrating the acceleration to get the velocity:

v(t) = ∫a(t)dt = ∫i + 2j + 2tk dt
= ti + 2tj + tk + C

Since the initial velocity is zero, we can solve for C:

v(0) = 0i + 0j + 0k + C = C

So C = (0,0,0) and the velocity becomes:

v(t) = ti + 2tj + tk

Now we can integrate the velocity to get the position:

r(t) = ∫v(t)dt = ∫ti + 2tj + tk dt
= 1/2 ti^2 + tj^2 + 1/2 tk^2 + C'

Using the initial position at t=0, we can solve for C':

r(0) = 1/2 (0)i^2 + (0)j^2 + 1/2 (0)k^2 + C' = (1,0,1)

So C' = (1,0,1) and the position becomes:

r(t) = 1/2 ti^2 + tj^2 + 1/2 tk^2 + (1,0,1)

Therefore, the position vector of the particle at any time t is:

r(t) = (1/2 t, 2t, 1/2 t) + (1,0,1)
= (1/2 t + 1, 2t, 1/2 t + 1)

Answer 2

isnt the integration be t i + 2t j + t^2 k + C

Answer 3

Yes Bill

Answer 4

You are correct, I apologize for the mistake in my previous response. The correct integration of the acceleration vector is:

v(t) = ∫a(t)dt = ∫(i + 2j + 2tk) dt
= ti + 2tj + (1/3)t^3k + C

Using the initial velocity of zero, we can solve for C:

v(0) = 0i + 0j + 0k + C = C

So C = (0,0,0) and the velocity vector becomes:

v(t) = ti + 2tj + (1/3)t^3k

Now, we can integrate the velocity to get the position:

r(t) = ∫v(t)dt = ∫(ti + 2tj + (1/3)t^3k)dt
= 1/2 ti^2 + tj^2 + (1/12)t^4k + C'

Using the initial position at t=0, we can solve for C':

r(0) = 1/2 (0)i^2 + (0)j^2 + (1/12)(0)^4k + C' = (1,0,1)

So C' = (1,0,1) and the position vector becomes:

r(t) = 1/2 ti^2 + tj^2 + (1/12)t^4k + (1,0,1)

Therefore, the position vector of the particle at any time t is:

r(t) = (1/2 t^2 + 1, 2t, 1/12 t^4 + 1)

Answer 5

v(t) = ∫a(t)dt = ∫(i + 2j + 2tk) dt
= t i + 2t j + t^2 k + C
!!!!!!

Answer 6

Yes, that is correct. Thank you for catching my mistake. The correct integration of a(t) is:

v(t) = ∫a(t)dt = ∫(i + 2j + 2tk) dt
= ti + 2tj + t^2k + C

Using the initial velocity of zero, we can solve for C:

v(0) = 0i + 0j + 0k + C = C

So C = (0,0,0) and the velocity vector becomes:

v(t) = ti + 2tj + t^2k

From there, we can integrate to find the position vector as shown in my previous response.