2.1) (i) W = mg downwards (ii) N = R = Normal Reaction from the ground upwards (iii) Fe = Force of engine towards the right (iv) f = friction towards the left (v) ma = Constant acceleration towards right. 2.2.1) v = 25 m/s u = 0 m/s ∆v = v - u = (25 - 0) m/s = 25 m/s x = X ∆t = 50 s [tex]a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} } [/tex] a = 0.5 m/s². 2.2.2) F = ma = 900 kg × 0.5 m/s² = 450 N. 2.2.3) [tex]2ax \: = \: {v}^{2} \: - \: {u}^{2} [/tex] [tex]x \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{2a} \: = \: \frac{{(25 \: \frac{m}{s})}^{2} \: - \: {(0 \: \frac{m}{s} )}^{2} }{2 \: \times \: 0.5 \: \frac{m}{ {s}^{2} } } \: = \: 625 \: m[/tex] 2.3) Fe = f + ma Fe - f = ma For velocity to be constant, a should be 0, or, a = 0, Fe = f = 270 N 2.4.1) v = 0 u = 25 m/s a = -0.5 m/s² v = u + at t = -u/a = -(25)/(-0.5) = 50 s. 2.4.2) x = -625/(2×(-0.5)) = 625 m.
