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.