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Question 1:An athlete runs in a straight line along a flat surface. He starts from rest and for 20 seconds accelerate at a constant rate. In this first 20 seconds he covers a distance of 100m. For the next 10 seconds he runs at a constant speed andthen decelerates at a constant rate for 5 seconds until he stops.a) What is the total distance that he ran?

Respuesta :

First, let's find the acceleration and the final velocity in the first 20 seconds, using the formula below:

[tex]\begin{gathered} d=V_0t+\frac{at^2}{2} \\ 100=0+a\cdot\frac{20^2}{2} \\ 100=a\cdot200 \\ a=0.5\text{ m/s}^2 \\ \\ V=V_0+a\cdot t \\ V=0+0.5\cdot20 \\ V=10\text{ m/s} \end{gathered}[/tex]

During the 10 seconds with constant speed, the distance he ran is:

[tex]\begin{gathered} d=v\cdot t \\ d=10\cdot10=100\text{ m} \end{gathered}[/tex]

Now, in the last 5 seconds, the distance is:

[tex]\begin{gathered} V=V_0+a\cdot t \\ 0=10+a\cdot5 \\ 5a=-10 \\ a=-2\text{ m/s} \\ \\ d=V_0t+\frac{at^2}{2} \\ d=10\cdot5-\frac{2\cdot5^2}{2} \\ d=50-25 \\ d=25\text{ m} \end{gathered}[/tex]

Therefore the total distance is:

[tex]d_{\text{total}}=100+100+25=225\text{ m}[/tex]

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