The acceleration is a bus is given by ax(t)=alpha*t, where alpha=1.23m/s^3 is a

Question

The acceleration is a bus is given by ax(t)=alpha*t, where alpha=1.23m/s^3 is a constant.
A) If the bus's velocity at the time t1 = 1.11s is 4.99 m/s, what is its velocity at time t2 = 2.11s?
(I know the answer to this is 6.97m/s but I don't know how to get it. I know you're supposed to integrate again to get the answer to part B but I don't even know the equation. My main concern is part B's answer.)

B) If the bus's position at time t1 = 1.11s is 6.08 m, what is its position at time t2 = 2.11s?

Explanation / Answer

b)

a = t

v = 0.5 t^2 + v0

>>>> 4.99 = 0.5*1.23*1.11*1.11+ v0

>>>> v0 = 4.2323 m/s

>>>> v = 0.5 t^2 + 4.2323

>>>> x = 0.5/3**t^3 + 4.2323 t + x0

>>>> 6.08 = 0.5/3*1.23*1.11*1.11*1.11 + 4.2323*1.11 + x0

>>>> x0 = 1.10178 m

>>>> x = 0.5/3**t^3 + 4.2323 t + 1.10178

>>>> x = 0.5/3*1.23*2.11*2.11*2.11 + 4.2323*2.11 + 1.10178

>>>> x = 12.0 m

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