Answer: The magnitude of his acceleration is \(1.033 m/s^2\).
Explanation:
Final velocity of the player= v = 3.1 m/s
Initial velocity of the player = u = 0 m/s
Time taken by the player = 3 s
Acceleration of the soccer player = a
Using first equation of the motion:
\(v=u+at\)
\(a=\frac{v-u}{t}=\frac{3.1- m/s-0 m/s}{3 s}=1.033 m/s^2\)
The magnitude of his acceleration is \(1.033 m/s^2\).
