1. **Problem Statement:** We are given the acceleration of a lorry as a function of time $t$ seconds from rest: $$a = \frac{3}{2}t - \frac{1}{10}t = \left(\frac{3}{2} - \frac{1}{10}\right) t = \frac{15}{10}t - \frac{1}{10}t = \frac{14}{10}t = 1.4t \text{ m/s}^2$$ We need to find: (a) The velocity after 10 seconds. (b) The distance traveled at time $t$ seconds. 2. **Formula and Rules:** - Acceleration $a(t)$ is the derivative of velocity $v(t)$ with respect to time: $$a(t) = \frac{dv}{dt}$$ - Velocity $v(t)$ is the derivative of displacement $s(t)$ with respect to time: $$v(t) = \frac{ds}{dt}$$ - To find velocity from acceleration, integrate acceleration with respect to time. - To find displacement from velocity, integrate velocity with respect to time. - Initial velocity and displacement are zero since the lorry starts from rest and from the origin. 3. **Find velocity $v(t)$:** $$v(t) = \int a(t) dt = \int 1.4t dt = 1.4 \int t dt = 1.4 \cdot \frac{t^2}{2} + C = 0.7 t^2 + C$$ Since the lorry starts from rest, $v(0) = 0$, so $C=0$. Thus, $$v(t) = 0.7 t^2$$ 4. **Calculate velocity after 10 seconds:** $$v(10) = 0.7 \times 10^2 = 0.7 \times 100 = 70 \text{ m/s}$$ 5. **Find displacement $s(t)$:** $$s(t) = \int v(t) dt = \int 0.7 t^2 dt = 0.7 \int t^2 dt = 0.7 \cdot \frac{t^3}{3} + C = \frac{0.7}{3} t^3 + C$$ Since the lorry starts from the origin, $s(0) = 0$, so $C=0$. Thus, $$s(t) = \frac{0.7}{3} t^3 = \frac{7}{30} t^3$$ **Final answers:** - (a) Velocity after 10 seconds: $$70 \text{ m/s}$$ - (b) Distance at time $t$ seconds: $$s(t) = \frac{7}{30} t^3 \text{ meters}$$