Question

a particle is moving with acceleration

(a(t) = 36t + 16),

inches per square second, where (t) is in seconds.

its position at time (t = 0) is (s(0) = 7) inches and its velocity at time (t = 0) is (v(0) = 6) inches per second.

i. what is the particular position function, (s(t))?

(s(t) =)

ii. what is the particles position at time (t = 11)?

(s(11) =) inches

Explanation:

Integrate acceleration to find velocity

$$ v(t) = \int a(t) \, dt = \int (36t + 16) \, dt = 18t^2 + 16t + C_1 $$

$$ v(0) = 6 \implies 18(0)^2 + 16(0) + C_1 = 6 \implies C_1 = 6 $$

$$ v(t) = 18t^2 + 16t + 6 $$

Integrate velocity to find position

$$ s(t) = \int v(t) \, dt = \int (18t^2 + 16t + 6) \, dt = 6t^3 + 8t^2 + 6t + C_2 $$

$$ s(0) = 7 \implies 6(0)^3 + 8(0)^2 + 6(0) + C_2 = 7 \implies C_2 = 7 $$

$$ s(t) = 6t^3 + 8t^2 + 6t + 7 $$

Evaluate position at t = 11

$$ s(11) = 6(11)^3 + 8(11)^2 + 6(11) + 7 $$

$$ s(11) = 6(1331) + 8(121) + 66 + 7 = 7986 + 968 + 73 = 9027 $$

Answer:

Question I

\(s(t) =\) <blank>\(6t^3 + 8t^2 + 6t + 7\)</blank>

Question II

\(s(11) =\) <blank>9027</blank> inches