A particle is moving with the given data. Find the position of the particle.
a(t) = 2t + 3, s(0) = 6, v(0) = −9
\(\text{We start with the acceleration to find the velocity}\\ v(t) = \displaystyle \int_0^t a(\tau) d\tau + v_0 = t^2+3t-9\\~\\ s(t) = \displaystyle \int_0^t v(\tau)d\tau + s_0 = \dfrac{t^3}{3}+\dfrac 3 2 t^2 - 9t + 6\)
thanks a lot :)
