Here, m = 0.40 kg, u = 10 m s–1 due north F = – 8.0 N (minus sign for opposite direction of force) $a = \frac{F}{m} = \frac{-8.0}{0.40} = -20 \text{ m}\,\text{s}^{-2}$ for $0 \leq t \leq 30 \text{ s}$ (i) At t = – 5 s, x = ut = 10 × (–5) = – 50 m (ii) At $t = 25 \text{ s},$ $x = u t + \frac{1}{2} a t^{2}$ $= 10 \times 25 + \frac{1}{2}(-20)(25)^{2}$ $= -6000 \text{ m}$ (iii) At t = 100 s, The problem is divided into two parts. First consider motion up to 30 s $x_{1} = u t + \frac{1}{2} a t^{2}$ $= 10 \times 30 + \frac{1}{2}(-20)(30)^{2}$ $= -8700 \text{ m}$ At t = 30 s, v = u + at = 10 – 20 × 30 = – 590 m s$^{-1}$ For motion 30 s to 100 s $x^{2} = v t$$= -590 \times 70 = -41300$ m $\therefore x = x_1 + x_2$$= -8700 - 41300$$= -50000 \text{ m} = -50 \text{ km}$