Given:

a is a constant
To find:
(a) v(t) in terms of v(0), and a
(b) x(t) in terms of x(0), v(0) and a
Step-by-step explanation:
(a) We can write,

Integrating both sides we can write,

Hence, the velocity is,

(b)
We can also write,

Integrating both sides we get,
![\begin{gathered} \int_(x(0))^(x(t))dx=\int_0^tv(t)dt \\ \int_(x(0))^(x(t))dx=\int_0^t[v(0)+at]dt \\ x(t)-x(0)=v(0)t+(1)/(2)at^2 \\ x(t)=x(0)+v(0)t+(1)/(2)at^2 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/5ndfalqkjskj31ul6e2q83zexkvsbxcqa4.png)
Hence,
