An athlete, initially at rest, takes 2 s to reach his maximum speed of `"36 km h"^(-1)`. The magnitude of his average accleration (in`ms^(-2)`) is

Answer

To solve the problem of finding the average acceleration of the athlete, we can follow these steps:

### Step 1: Convert the final velocity from km/h to m/s
The final speed of the athlete is given as 36 km/h. We need to convert this to meters per second (m/s) using the conversion factor:
\[ 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s} \]
Thus,
\[ 36 \text{ km/h} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s} \]

### Step 2: Identify the initial and final velocities
The athlete starts from rest, so the initial velocity (\(V_i\)) is:
\[ V_i = 0 \text{ m/s} \]
The final velocity (\(V_f\)) after 2 seconds is:
\[ V_f = 10 \text{ m/s} \]

### Step 3: Calculate the change in velocity
The change in velocity (\(\Delta V\)) is given by:
\[ \Delta V = V_f - V_i = 10 \text{ m/s} - 0 \text{ m/s} = 10 \text{ m/s} \]

### Step 4: Calculate the average acceleration
Average acceleration (\(a\)) is defined as the change in velocity divided by the time taken:
\[ a = \frac{\Delta V}{\Delta t} \]
Where \(\Delta t\) is the time taken, which is 2 seconds. Therefore,
\[ a = \frac{10 \text{ m/s}}{2 \text{ s}} = 5 \text{ m/s}^2 \]

### Final Answer
The magnitude of the average acceleration is:
\[ \boxed{5 \text{ m/s}^2} \]

---