Question

An athlete start from rest with an acceleration 2 metre per second square for sometime and then Declerate with 3 metre per second square and finally came to rest. if total length of track is 100 M then maximum speed attained by athletes

Answer 1

600 is the answer!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

Answer:

15.5 m/s

Explanation:

Case I:

The athlete starts from rest to reach its maximum velocity v with an acceleration of $2\ m/s^2$.

Let us find out the distance traveled by the athlete in this interval.

$v^2=u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{v^2-0^2}{2\times 2}\\\Rightarrow s = \dfrac{v^2}{4}\,\,\,\,\,\, (1)$

Case II:

Now the athlete on reaching this maximum velocity v, decelerates at a constant rate to come to rest.

Let us find out the distance traveled by the athlete in this interval of time.

$v^2=u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{0^2-v^2}{2\times (-3)}\\\Rightarrow s = \dfrac{v^2}{6}\,\,\,\,\,\, (2)$

Now summing up the distance traveled in both the cases, we have the total distance 100 m.

$\therefore \dfrac{v^2}{4}+\dfrac{v^2}{6}=100\\\Rightarrow v^2(\dfrac{1}{4}+\dfrac{1}{6})=100\\\Rightarrow v^2(\dfrac{10}{24})=100\\\Rightarrow v^2=\dfrac{24}{10}\times 100\\\Rightarrow v^2=240$

Taking square root on both the sides, we have

$v=\pm \sqrt{240}\\v = \pm 15.5\ m/s$

Hence, the magnitude of the maximum speed attained by the athlete is 15.5 m/s.