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 answer:

F =  5402.0979 N

Step-by-step explanation:

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

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