--- Integral Variable Acceleration Topic Assessment Answers -
(b) ( s(t) = \int (\sin 2t + \cos t) dt = -\frac{1}{2}\cos 2t + \sin t + D ) ( s(0) = -\frac12(1) + 0 + D = -\frac12 + D = 0 \Rightarrow D = \frac12 ) [ s(t) = -\frac12\cos 2t + \sin t + \frac12 ] (a) ( v(t) = \int (12t^2 - 8t + 2) dt = 4t^3 - 4t^2 + 2t + C ) ( v(1) = 4 - 4 + 2 + C = 2 + C = 5 \Rightarrow C = 3 ) [ v(t) = 4t^3 - 4t^2 + 2t + 3 ]
(b) ( s(t) = \int \left(6 - \frac{4}{t+1}\right) dt = 6t - 4\ln(t+1) + D ) ( s(0) = 0 - 0 + D = 0 \Rightarrow D = 0 ) [ s(t) = 6t - 4\ln(t+1) ] (a) ( v(t) = \int 12 t^{1/2} dt = 12 \cdot \frac{2}{3} t^{3/2} + C = 8 t^{3/2} + C ) ( v(4) = 8 \cdot 8 + C = 64 + C = 10 \Rightarrow C = -54 ) [ v(t) = 8t^{3/2} - 54 ] --- Integral Variable Acceleration Topic Assessment Answers
(a) Find ( v(t) ) (3 marks) (b) Find ( s(t) ) (3 marks) (c) Calculate the total distance travelled between ( t = 1 ) and ( t = 4 ) seconds, explaining how you treat any change of direction. (3 marks) Q1 (a) ( v(t) = \int (6t - 4), dt = 3t^2 - 4t + C ) ( v(0) = 5 \Rightarrow C = 5 ) [ v(t) = 3t^2 - 4t + 5 ] (b) ( s(t) = \int (\sin 2t +
Distance ( = s(4) - s(1) ) ( s(4) = 256 - \frac{256}{3} + 16 + 12 - \frac{2}{3} ) ( = 284 - \frac{258}{3} = 284 - 86 = 198 ) ( s(1) = 3 ) (given) [ \text{Distance} = 198 - 3 = 195 \ \text{m} ] --- Integral Variable Acceleration Topic Assessment Answers
