## Essential University Physics: Volume 1 (4th Edition) Clone

We know that $v=\int_0^t a dt$ $v=\int_0^t(a_{\circ}+bt)dt$ $v=v_{\circ}+a_{\circ}t+(\frac{b}{2})t^2$ (b) $x=\int _0^t v dt$ $x=\int _0^t (v_{\circ}+a_{\circ}t+\frac{b}{2}t^2)dt$ $x=x_{\circ}+v_{\circ}t+(\frac{a_{\circ}}{2})t^2+(\frac{b}{6})t^3$