Answer:
(13/3, 0)
Step-by-step explanation:
a(t) = t i + 2 j
Use integration to find v(t).
v(t) = ∫ a(t) dt
v(t) = ∫ (t i + 2 j) dt
v(t) = (½ t² + C₁) i + (2t + C₂) j
Since the initial velocity is 0, v(0) = 0 i + 0 j:
0 i + 0 j = (½ (0)² + C₁) i + (2 (0) + C₂) j
C₁ = 0 and C₂ = 0
v(t) = ½ t² i + 2t j
Use integration to find s(t).
s(t) = ∫ v(t) dt
s(t) = ∫ (½ t² i + 2t j) dt
s(t) = (⅙ t³ + C₃) i + (t² + C₄) j
The particle starts at (3, -4), so s(0) = 3 i − 4 j:
3 i − 4 j = (⅙ (0)³ + C₃) i + ((0)² + C₄) j
C₃ = 3 and C₄ = -4
s(t) = (⅙ t³ + 3) i + (t² − 4) j
After 2 seconds, the position is:
s(2) = (⅙ (2)³ + 3) i + ((2)² − 4) j
s(2) = 13/3 i + 0 j
The coordinates of the point is (13/3, 0).