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Given ( a(t) = \fracdvdt = 6t + 4 ). Integrate: [ v(t) = \int (6t + 4) , dt = 3t^2 + 4t + C_1 ] Using ( v(0)=5 ): ( 5 = 0 + 0 + C_1 \implies C_1 = 5 ). Thus, ( v(t) = 3t^2 + 4t + 5 ).
[ v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 \ \textm/s ] [ a(2) = 6(2) - 12 = 0 \ \textm/s^2 ]
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.
Find when ( v(t)=0 ): ( 2t-4=0 \implies t=2 ) s.
