Part II: s(t), v(t) and a(t) : position vector, velocity vector and acceleration vector related Def: Instantaneous velocity (speed) (1) V(t) = s'(t) =- Instantaneous acceleration: (2) a(t) = v '(t) = at du (3) a(t) = s"(t) Thus: (4) s(t) = fv(t) dt v(t) = Sa(t) dt Your Problem: is to prove (4) i) Given a(t) = -g g = gravity downward Find an expression for v(t) V(t) = ii) Given an initial condition of : v (0) = vo, solve for C and find general equation for v(t): iii) With your expression for v(t), find s(t). Then use: s(0) = so to find constant C and general equation for s(t): S(t) =Problem [3] (1) An object is moving with velocity v(t) = 12 + 4t-5. Find the displacement and the total distance travelled from t[0, 7] Displacement: s(t) Total Distance = S,, | v(t) | dt = S, v(t)dt + S - v(t) dt