Acceleration 7147

The runner ran the track 100m in 10.2 seconds. He ran the first 20m with a uniformly accelerated movement. Then he ran evenly. What was its acceleration on the 20m section, and what was its maximum speed?

Correct answer:

a =  3.4602 m/s²
v =  11.7647 m/s

Step-by-step explanation:

$$\begin{gathered} s=100\text{m} \\ {s}_2=20\text{m} \\ t=10.2\text{s} \\ {s}_2=\frac{1}{2}\cdot a\cdot t_1^2 \\ v=a\cdot t_1 \\ s-s_2=v\cdot (t-t_1) \\ 2\cdot s_2=a\cdot t_1^2=v\cdot t_1 \\ s-s_2=v\cdot (t-t_1) \\ s-s_2=tv-v{t}_1=t\cdot v-2\cdot s_2 \\ v=\frac{s+s_2}{t}=\frac{100+20}{10.2}=\frac{200}{17}\doteq 11.7647\text{m/s} \\ {t}_1=2\cdot s_2/v=2\cdot s_2/\frac{200}{17}=2\cdot 20/11.7647=\frac{17}{5}=3.4\text{s} \\ a=v/t_1=\frac{200}{17}/\frac{17}{5}=\frac{200}{17}\cdot \frac{5}{17}=\frac{200\cdot 5}{17\cdot 17}=\frac{1000}{289}=\frac{1000}{289}{\text{m/s}}^2=3.4602{\text{m/s}}^2 \end{gathered}$$

$$\begin{gathered} v=\frac{200}{17}\doteq 11.7647\text{m/s} \\ {v}_2=v\to \text{km/h}=v\cdot 3.6\text{km/h}=11.7647\cdot 3.6\text{km/h}=42.353\text{km/h} \end{gathered}$$

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