Hello, can someone help me. thank you sir.

Question

An object moves in space with acceleration a(t)=6ti+(12t+2)j+e^tk. When t=0, the object is at (0,3,5) with initial velocity v(0)=2i+k. Find the position vector, r(t), of the object.

Answer 1

v(t)=∫a(t)dt

v(t)=(3t²+c₁)i+(6t²+2t+c₂)k+(e^t+c₃)k

From initial conditions, c₁=2, c₂=c₃=0

v(t)=(3t²+2)i+(6t²+2t)k+(e^t)k

r(t)=(t³+2t+c₄)i+(2t³+t²+c₅)k+(e^t+c₆)k

From initial conditions, c₄=0, c₅=3 and c₆=4

r(t)=(t³+2t)i+(2t³+t²+3)k+(e^t+4)k

Answer 2

Velocity v(t) = ∫a(t)dt = 3t²i + (6t² + 2t)j + e^tk + C; v(0) = k + C = 2i + k; C = 2i; v(t) = (3t²+ 2)i + (6t² + 2t)j + e^tk.

Position s(t) = ∫v(t)dt = (t³ + 2t)i + (2t³ + t²)j + e^tk + D; s(0) = k + D = <0,3,5>; D = <0,3,4> = 3j + 4k;

s(t) = (t³ + 2t)i + (2t³ + t² + 3)j + (e^t + 4)k