v = 2t – 1c Find the distance moved from = 1 to = 4 t t x = t2 + e 1 t + c – Antiderivative sub t = 4 x4 = 16 + e 3 + c – correct. sub t = 1 x1 = 1 + e0 + c = 2 + c Correct x4 x1 = 14 + e 3 – – ≈14.05 metres integration with correct solution. dm = 1d dt = 0 and at hr, t m = 8g t = 2 m = 1 g. = Correct diff equ with = kt + c k and c found = kt + c = u Sub t = 0, m = 8 6 = c Sub t = 2, m = 1 3 = 2k + 6 so k = 9 – 2 4 = + 6 Correct time = r If m = 0 t = 24 = hours 9
1c v = 2t – Find the distance moved from t = 1 to t = 4 x = t2 + e 1 – t + c sub t = 4 x4 = 16 + e – 3 + c sub t = 1 x1 = 1 + e0 + c = 2 + c x4 – x1 = 14 + e – 3 ≈14.05 metres Antiderivative correct. Correct integration with correct solution. 1d dm = dt t = 0 m = 8g and at t = 2 hr, m = 1 g. = = kt + c = kt + c Sub t = 0, m = 8 6 = c Sub t = 2, m = 1 3 = 2k + 6 so k = – 9 2 4 = + 6 If m = 0 t = 24 = hours 9 Correct diff equ with k and c found = u Correct time = r
4b_practice_examination__number_4_schedule.docx