Accelerated motion - mechanics

The delivery truck with a total weight of 3.6 t accelerates from 76km/h to 130km/h in the 0.286 km long way. How much was the force needed to achieve this acceleration?

Correct result:

F =  5402.0979 N

Solution:

$$\begin{gathered} m=3.6\text{ t}\to \text{ kg}=3.6\cdot 1000\text{ kg}=3600\text{ kg } \\ s=0.286\text{ km}\to \text{ m}=0.286\cdot 1000\text{ m}=286\text{ m } \\ {v}_1=76\text{ km/h}\to \text{ m/s}=76\mathrm{/}3.6\text{ m/s}=21.11111\text{ m/s } \\ {v}_2=130\text{ km/h}\to \text{ m/s}=130\mathrm{/}3.6\text{ m/s}=36.11111\text{ m/s } \\ s=v_1\cdot t+{\displaystyle \frac{1}{2}}a{t}^2 \\ a={\displaystyle \frac{v_2-v_1}{t}} \\ s=v_1\cdot t+{\displaystyle \frac{1}{2}}{\displaystyle \frac{v_2-v_1}{t}}{t}^2 \\ s=v_1\cdot t+{\displaystyle \frac{1}{2}}(v_2-v_1)t \\ t={\displaystyle \frac{s}{v_1+(v_2-v_1)\mathrm{/}2}}={\displaystyle \frac{286}{21.1111+(36.1111-21.1111)\mathrm{/}2}}={\displaystyle \frac{5148}{515}}=9{\displaystyle \frac{513}{515}}\doteq 9.9961\text{ s } \\ a={\displaystyle \frac{v_2-v_1}{t}}={\displaystyle \frac{36.1111-21.1111}{9.9961}}\doteq 1.5006{\text{m/s}}^2 \\ {s}_1=v_1\cdot t+{\displaystyle \frac{1}{2}}\cdot a\cdot t^2=21.1111\cdot 9.9961+{\displaystyle \frac{1}{2}}\cdot 1.5006\cdot 9.9961^2=286\text{ m } \\ {s}_1=s \\ F=m\cdot a=3600\cdot 1.5006=5402.0979\text{ N} \end{gathered}$$

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