Aparticle starts from rest at the point (3,-4) with an acceleration vector of a(t) = t i + 2j
find the location of the particle after 2 seconds.

(13/3, 0)

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).

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2