Motion Problems And Solutions Mathalino Upd | Rectilinear

Displacement from t=2 to t=6: [ \int_2^6 (2t-4) dt = [t^2 - 4t]_2^6 = (36-24) - (4-8) = 12 - (-4) = 16 \ \textm ] Distance part 2 = ( 16 ) m (positive, no absolute needed).

[ v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 \ \textm/s ] [ a(2) = 6(2) - 12 = 0 \ \textm/s^2 ] rectilinear motion problems and solutions mathalino upd

(a) ( v=-3 \ \textm/s, a=0 ); (b) ( t=1,3 \ \texts ); (c) Displacement = 20 m, Distance = 28 m. Problem 2: Variable Acceleration (Integration Required) Statement: The acceleration of a particle in rectilinear motion is given by ( a(t) = 6t + 4 \ \textm/s^2 ). At ( t=0 ), the velocity ( v_0 = 5 \ \textm/s ) and position ( s_0 = 2 \ \textm ). Find the position function ( s(t) ). Displacement from t=2 to t=6: [ \int_2^6 (2t-4)

– 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. At ( t=0 ), the velocity ( v_0

Now, ( v(t) = \fracdsdt \implies s(t) = \int (3t^2 + 4t + 5) , dt = t^3 + 2t^2 + 5t + C_2 ). Using ( s(0)=2 ): ( 2 = 0 + 0 + 0 + C_2 \implies C_2 = 2 ).

Find when ( v(t)=0 ): ( 2t-4=0 \implies t=2 ) s.