Answer:
1.09 m/s²
Explanation:
Given:
Final speed of the sprinter (v) = 10.4 m/s
Initial speed of the sprinter (u) = 0 m/s (As the sprinter starts from rest position)
Time taken to reach the speed(Δt) = 9.58 s
Average acceleration is defined as the rate of change of velocity.
So, the acceleration of the sprinter is given as:
[tex]a=\frac{v-u}{\Delta t}[/tex]
Plug in all the given values and solve for acceleration, 'a'. This gives,
[tex]a=\frac{10.4-0}{9.58}\\\\a=1.09\ m/s^2[/tex]
Therefore, the magnitude of the average acceleration of the sprinter is 1.09 m/s².
