Respuesta :
Answer:
(a) The collision lasts for 0.053 s.
(b) The deceleration is 180.5 m/s².
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:
[tex]v^2=u^2+2aS[/tex]
Expressing in terms of 'a', we get:
[tex]a=\frac{v^2-u^2}{2S}[/tex]
Plug in the given values and solve for 'a'. This gives,
[tex]a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2[/tex]
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:
[tex]v=u+at\\\\t=\frac{v-u}{a}[/tex]
Plug in the given values and solve for 't'. This gives,
[tex]t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s[/tex]
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.
