Find the position vector of a particle whose acceleration vector is a = (6t, 2) with an initial velocity vector (0, 0) and initial position vector (3, 0)

Please I need your help with explanation. Thank you

this is just straightforward integration and evaluation at given points to determine the constant.

acceleration: a = (6t,2)

velocity: v = (3t^2,2t) + c
since v(0) = 0, c = (0,0), and thus v = (3t^2,2t)

position: s = (t^3,t^2) + c
since s(0) = (3,0), c = (3,0) and thus

s(t) = (t^3+3, t^2)

What else I need to do to get the answer, please?

huh? I gave you the position vector. Was there something else you wanted????

To find the position vector of a particle, we can integrate the acceleration vector twice with respect to time.

Given: acceleration vector a = (6t, 2) with an initial velocity vector v₀ = (0, 0) and initial position vector r₀ = (3, 0).

Step 1: Find the velocity vector.
We integrate the acceleration vector with respect to time to find the velocity vector.
∫(6t, 2) dt = (3t², 2t) + C₁
Using the initial velocity v₀ = (0, 0), we plug in t = 0 and solve for C₁:
(3(0)², 2(0)) + C₁ = (0, 0) + C₁
C₁ = (0, 0)

Therefore, the velocity vector v = (3t², 2t).

Step 2: Find the position vector.
We integrate the velocity vector with respect to time to find the position vector.
∫(3t², 2t) dt = (t³, t²) + C₂
Using the initial position r₀ = (3, 0), we plug in t = 0 and solve for C₂:
(0³, 0²) + C₂ = (3, 0) + C₂
C₂ = (3, 0)

Therefore, the position vector r = (t³ + 3, t²) + C₂.

The final position vector of the particle is r = (t³ + 3, t²) + (3, 0).
Simplifying this expression, we get:
r = (t³ + 3, t²)

So the position vector of the particle is r = (t³ + 3, t²), where t represents time.