Rectilinear Motion Problems And Solutions Mathalino Upd -

[ v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 \ \textm/s ] [ a(2) = 6(2) - 12 = 0 \ \textm/s^2 ]

Compute positions: [ s(0) = 2,\ s(1) = 1 - 6 + 9 + 2 = 6,\ s(3) = 27 - 54 + 27 + 2 = 2,\ s(5) = 125 - 150 + 45 + 2 = 22 ] Displacement = ( s(5) - s(0) = 22 - 2 = 20 ) m (positive, to the right). rectilinear motion problems and solutions mathalino upd

Displacement from t=0 to t=2: [ \int_0^2 (2t-4) dt = [t^2 - 4t]_0^2 = (4-8) - 0 = -4 \ \textm ] Distance part 1 = ( | -4 | = 4 ) m. [ v(2) = 3(4) - 12(2) + 9

– Need to account for direction changes at t=1 and t=3. From t=0 to 1: ( |s(1)-s(0)| = |6-2| = 4 ) m. From t=1 to 3: ( |s(3)-s(1)| = |2-6| = 4 ) m. From t=3 to 5: ( |s(5)-s(3)| = |22-2| = 20 ) m. Total distance = ( 4 + 4 + 20 = 28 ) m. From t=0 to 1: ( |s(1)-s(0)| = |6-2| = 4 ) m

Therefore, ( s(t) = t^3 + 2t^2 + 5t + 2 ) meters.

→ ( v(t)=0 ) [ 3t^2 - 12t + 9 = 0 \implies t^2 - 4t + 3 = 0 \implies (t-1)(t-3)=0 ] Thus, ( t = 1 ) s and ( t = 3 ) s.