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`        Suppose the coefficient of friction between an athlete's shoes and the ground is 0.9. The athlete starts running with an acceleration a. Then the equation will be ma=mμ. This equation means that the body is in equilibrium but according to the question the athlete is running. How is this possible?`
5 months ago

```							To reach the minimum time he has to move with maximum possible acceleration.Let the maximum acceleration be a.ma-μR=0ma=μmga=μg=0.9x10=9 m/s2A) Initial velocity=u= o m/st=?a=9m/s2 s=5mFrom s= ut +1/2 at²50=0+1/2 at²t=100/9t=10/3 sec=3.3 secb) AFTER MOVING 50m, velocity of the athelete given by :V=u+at=0+9x 10/3he has to stop in minimum time, so deceleration is -a=-9 m/s2 R=mgma=μRa=μg =9m/s2[deceleration]u1=30m/sv1=0 m/s]t=v1-u1/at=0-30/-at=-30/-9t=10/3 sec=3.3 sec
```
5 months ago
```							Hiii The force that the ground can exert on man is equal and opposite to the force of friction on the ground. There is misconceptions in this . That does not mean man should be at rest .
```
5 months ago
```							AFTER MOVING 50m, velocity of the athelete given by :V=u+at=0+9x 10/3he has to stop in minimum time, so deceleration is -a=-9 m/s2 R=mgma=μRa=μg =9m/s2[deceleration]u1=30m/sv1=0 m/s]t=v1-u1/at=0-30/-at=-30/-9t=10/3 sec=3.3 sec
```
5 months ago
```							Dear student,                         velocity of the athelete given by :V=u+at=0+9x 10/3he has to stop in minimum time, so deceleration is -a=-9 m/s2 R=mgma=μRa=μg =9m/s2[deceleration]u1=30m/sv1=0 m/s]t=v1-u1/at=0-30/-at=-30/-9t=10/3 sec=3.3 sec  hope this helps you
```
2 months ago
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