The car

The car weight 1280 kg, 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:

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

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