The car

The car weight 1280 kg and increased its speed from 7.3 m/s to 63 km/h on a track of 37.2 m. What force did the car engine have to exert?

Correct answer:

F =  4352 N

Step-by-step explanation:

$$\begin{gathered} m=1280\text{kg} \\ {v}_1=7.3\text{m/s} \\ {v}_2=63\text{km/h}\to \text{m/s}=63:3.6\text{m/s}=17.5\text{m/s} \\ s=37.2\text{m} \\ s=v_1\cdot t+\frac{1}{2}\cdot a\cdot t^2 \\ {v}_2=v_1+a\cdot t \\ {v}_2-v_1=at \\ s=v_1\cdot t+\frac{1}{2}\cdot (v_2-v_1)\cdot t \\ t=\frac{2\cdot s}{2\cdot v_1+v_2-v_1} \\ t=\frac{2\cdot s}{v_1+v_2}=\frac{2\cdot 37.2}{7.3+17.5}=3\text{s} \\ a=\frac{v_2-v_1}{t}=\frac{17.5-7.3}{3}=\frac{17}{5}=3.4{\text{m/s}}^2 \\ F=m\cdot a=1280\cdot \frac{17}{5}=\frac{1280\cdot 17}{5}=\frac{21760}{5}=4352=4352\text{N} \\ \text{ Verifying Solution: } \\ {s}_2=v_1\cdot t+\frac{1}{2}\cdot a\cdot t^2=7.3\cdot 3+\frac{1}{2}\cdot 3.4\cdot 3^2=\frac{186}{5}=37.2\text{m} \end{gathered}$$

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