--- Integral Variable Acceleration Topic Assessment Answers Apr 2026

(b) ( s(t) = \int (3t^2 - 4t + 5), dt = t^3 - 2t^2 + 5t + D ) ( s(0) = 2 \Rightarrow D = 2 ) [ s(t) = t^3 - 2t^2 + 5t + 2 ]

(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 ).

(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) A particle moves with acceleration [ a(t) = 12t^2 - 8t + 2 ]

At ( t = 1 ), ( v = 5 \ \text{m/s} ), ( s = 3 \ \text{m} ). (b) ( s(t) = \int (3t^2 - 4t

Find: (a) The velocity ( v(t) ) (2 marks) (b) The displacement ( s(t) ) (2 marks) (c) The displacement when ( t = 3 ) seconds (2 marks) The acceleration of a particle is given by [ a(t) = \frac{4}{(t+1)^2}, \quad t \ge 0 ] At ( t = 0 ), ( v = 2 \ \text{m/s} ) and ( s = 0 ).

(a) Find ( v(t) ) (3 marks) (b) Find ( s(t) ) (2 marks) A particle moves with acceleration [ a = 12\sqrt{t} \quad (t \ge 0) ] Given that ( v = 10 ) when ( t = 4 ) and ( s = 20 ) when ( t = 4 ): (a) Find ( v(t) ) (3 marks) (b)

(c) ( s(t) = \int v(t) dt = \frac{t^3}{2} - \frac{t^4}{24} + D ), ( s(0) = 0 \Rightarrow D = 0 ) Distance ( = s(9) = \frac{729}{2} - \frac{6561}{24} ) ( = 364.5 - 273.375 = 91.125 \ \text{m} ) (or ( \frac{729}{8} \ \text{m} )) (a) ( v(t) = \int (2\cos 2t - \sin t) dt = \sin 2t + \cos t + C ) ( v(0) = 0 + 1 + C = 1 \Rightarrow C = 0 ) [ v(t) = \sin 2t + \cos t ]

(c) ( s(3) = 27 - 18 + 15 + 2 = 26 \ \text{m} ) (a) ( v(t) = \int 4(t+1)^{-2} dt = -4(t+1)^{-1} + C ) ( v(0) = -4 + C = 2 \Rightarrow C = 6 ) [ v(t) = 6 - \frac{4}{t+1} ]

(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 ))

Avatar Mobile
Main Menu x
X
×