A cyclist acceleration from rest to a velocity of `72 km hr^(-1)` in 10sec. If the cyclist is in straight track the acceleration of the cyclist is:

Options

A. `7.2 m s^(-2)`
B. `120 ms^(-2)`
C. `2 ms^(-2)`
D. `0.2 m s^(-2)`

Explanation

To find the acceleration of the cyclist, we will follow these steps:

### Step 1: Convert the final velocity from km/h to m/s
The final velocity is given as \( 72 \, \text{km/h} \). To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \]
Thus,
\[ 72 \, \text{km/h} = 72 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 72 \times \frac{5}{18} = 20 \, \text{m/s} \]

### Step 2: Identify the initial velocity
The cyclist starts from rest, so the initial velocity \( u \) is:
\[ u = 0 \, \text{m/s} \]

### Step 3: Use the formula for acceleration
We will use the equation of motion that relates final velocity, initial velocity, acceleration, and time:
\[ v = u + at \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time taken.

### Step 4: Substitute the known values into the equation
We know:
- \( v = 20 \, \text{m/s} \)
- \( u = 0 \, \text{m/s} \)
- \( t = 10 \, \text{s} \)

Substituting these values into the equation:
\[ 20 = 0 + a \times 10 \]

### Step 5: Solve for acceleration
Rearranging the equation gives:
\[ 20 = 10a \]
Dividing both sides by 10:
\[ a = \frac{20}{10} = 2 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the cyclist is \( 2 \, \text{m/s}^2 \).

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