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To find the velocity of the bus at time t = 2s, we need to integrate the acceleration function with respect to time.

Given that the acceleration is given by a = (3 m/s³) t, we can integrate it to find the velocity function.

∫ a dt = ∫ (3 m/s³) t dt

Integrating both sides:

v = (3/2 m/s³) t² + C

To determine the constant of integration (C), we can use the given information that the velocity at t = 1s is 22 m/s.

v(1) = (3/2 m/s³) (1)² + C = 22 m/s

Simplifying:

(3/2) + C = 22

C = 22 - (3/2)

C = 22 - 1.5

C = 20.5

Therefore, the velocity function is:

v = (3/2 m/s³) t² + 20.5

Now we can calculate the velocity at t = 2s:

v(2) = (3/2 m/s³) (2)² + 20.5

v(2) = (3/2 m/s³) (4) + 20.5

v(2) = 6 m/s² + 20.5

v(2) = 26.5 m/s

So, the velocity of the bus at time t = 2s is 26.5 m/s.

To find the position of the bus at time t = 2s, we need to integrate the velocity function with respect to time.

∫ v dt = ∫ [(3/2 m/s³) t² + 20.5] dt

Integrating both sides:

s = (1/2) (3/2 m/s³) t³ + 20.5t + D

To determine the constant of integration (D), we can use the given information that the position at t = 1s is 30m.

s(1) = (1/2) (3/2 m/s³) (1)³ + 20.5(1) + D = 30m

Simplifying:

(9/4) + 20.5 + D = 30

(9/4) + 20.5 + D = 120/4

(9/4) + 20.5 + D = 30

D = (120/4) - (9/4) - 20.5

D = 30 - 9/4 - 20.5

D = 30 - 2.25 - 20.5

D = 7.25 - 20.5

D = -13.25

Therefore, the position function is:

s = (1/2) (3/2 m/s³) t³ + 20.5t - 13.25

Now we can calculate the position at t = 2s:

s(2) = (1/2) (3/2 m/s³) (2)³ + 20.5(2) - 13.25

s(2) = (1/2) (3/2 m/s³) (8) + 41 - 13.25

s(2) = 12 m + 41 - 13.25

s(2) = 53.75 m

So, the position of the bus at time t = 2s is 53.75 meters.

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