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.