Deceleration of car

The car has a speed of 60 km/h and, after a 100 m journey speed of 40 km/h. What is the deceleration of a car if we assume that its movement is constantly slowed down?

Correct answer:

a =  0.7716 m/s²

Step-by-step explanation:

$$\begin{gathered} v_1=60\text{km/h}\to \text{m/s}=60:3.6\text{m/s}=16.66667\text{m/s} \\ s=100\text{m} \\ {v}_2=40\text{km/h}\to \text{m/s}=40:3.6\text{m/s}=11.11111\text{m/s} \\ s=v_1\cdot t-\frac{1}{2}\cdot a\cdot t^2 \\ a=\frac{v_1-v_2}{t} \\ s=v_1\cdot t-0.5\cdot (v_1-v_2)/t\cdot t^2 \\ s=v_1\cdot t-0.5\cdot (v_1-v_2)\cdot t \\ 100=16.666666666667\cdot t-0.5\cdot (16.666666666667-11.111111111111)\cdot t \\ 13.888889t=100 \\ t=\frac{100}{13.88888889}=7.2 \\ t=\frac{36}{5}\doteq 7.2 \\ a=\frac{v_1-v_2}{t}=\frac{16.6667-11.1111}{7.2}=\frac{125}{162}\doteq 0.7716{\text{m/s}}^2 \\ \text{ Verifying Solution: } \\ {s}_1=v_1\cdot t-\frac{1}{2}\cdot a\cdot t^2=16.6667\cdot 7.2-\frac{1}{2}\cdot 0.7716\cdot 7.2^2=100\text{m} \\ s=s_1 \\ {v}_3=v_1-a\cdot t=16.6667-0.7716\cdot 7.2=\frac{100}{9}\doteq 11.1111\text{m/s} \\ {v}_3=v_2 \end{gathered}$$

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