Answer:
(a) The collision lasts for 0.053 s.
(b) The deceleration is 180.5 m/s².
Step-by-step explanation:
Given:
Initial velocity of the player (u) = 9.50 m/s
Final velocity of the player (v) = 0 m/s (Comes to a stop)
Displacement of the player (S) = 0.250 m
We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:
![v^2=u^2+2aS](https://img.qammunity.org/2021/formulas/physics/college/kj5iy82gakuk3f0ujfnv5oddqzuytz7vhd.png)
Expressing in terms of 'a', we get:
![a=(v^2-u^2)/(2S)](https://img.qammunity.org/2021/formulas/physics/college/aiyzve9bnhi2644ezzicqcaeigky4ocswp.png)
Plug in the given values and solve for 'a'. This gives,
![a=(0-9.50^2)/(2* 0.250)\\\\a=(-90.25)/(0.5)=-180.5\ m/s^2](https://img.qammunity.org/2021/formulas/physics/college/c5kbb8o0ujrwlz3fin8aqv4v16givl1ozb.png)
Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².
Now, using the first equation of motion, we have:
![v=u+at\\\\t=(v-u)/(a)](https://img.qammunity.org/2021/formulas/physics/college/bh7o48jll0z8l6la2spt6l13kaqfi7fpv8.png)
Plug in the given values and solve for 't'. This gives,
![t=(0-9.5)/(-180.5)\\\\t=0.053\ s](https://img.qammunity.org/2021/formulas/physics/college/whg991al4s57wf309rnmj7vehn33z2lk1l.png)
Therefore, the the collision will last for 0.053 s.
(a) The collision lasts for 0.053 s.
(b) The deceleration is 180.5 m/s².