--- Integral Variable Acceleration Topic Assessment Answers -

(a) Find the velocity function ( v(t) ) (2 marks) (b) Find the time when the car is momentarily at rest again (2 marks) (c) Find the distance travelled up to that time (1 mark) A particle’s acceleration is given by [ a(t) = 2\cos(2t) - \sin t ] At ( t = 0 ), ( v = 1 ), ( s = 0 ).

(c) Check if ( v(t) = 0 ) in [1,4]: ( v(t) = 4t^3 - 4t^2 + 2t + 3 ) Test ( t=1 ): ( 4 - 4 + 2 + 3 = 5 >0 ) Test ( t=0 ): ( 3 >0 ), cubic positive, likely no root. Check derivative: ( 12t^2-8t+2>0 ) (discriminant 64-96<0) so ( v(t) ) increasing, always positive. No change of direction. --- Integral Variable Acceleration Topic Assessment Answers

(b) ( v(t) = 0 \Rightarrow \frac{t^2}{2}\left(3 - \frac{t}{3}\right) = 0 ) ( t = 0 ) or ( t = 9 ) seconds (answer: ( t = 9 )) (a) Find the velocity function ( v(t) )

(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 ] No change of direction

(a) Find ( v(t) ) (3 marks) (b) Find ( s(t) ) (3 marks) A particle moves with acceleration [ a(t) = 12t^2 - 8t + 2 ]

(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 ]