If a car at rest, accelerates uniformly to a speed of `144 km//h` in `20s`, it covers a distance of

Options

A. `2880m`
B. `1440 m`
C. `400 m`
D. `20m`

Answer

To solve the problem step by step, we will follow the physics principles of motion in one dimension.

### Step 1: Convert the final speed from km/h to m/s
The final speed of the car is given as \( 144 \, \text{km/h} \). We need to convert this to meters per second (m/s).

\[ \text{Final speed} (v) = 144 \, \text{km/h} = \frac{144 \times 1000 \, \text{m}}{3600 \, \text{s}} = 40 \, \text{m/s} \]

### Step 2: Identify the initial speed and time
The car starts from rest, so the initial speed \( u = 0 \, \text{m/s} \). The time taken to reach the final speed is \( t = 20 \, \text{s} \).

### Step 3: Calculate the acceleration
Using the formula for acceleration \( a \):

\[ a = \frac{v - u}{t} \]

Substituting the known values:

\[ a = \frac{40 \, \text{m/s} - 0 \, \text{m/s}}{20 \, \text{s}} = \frac{40}{20} = 2 \, \text{m/s}^2 \]

### Step 4: Calculate the distance covered
We can use the formula for distance \( s \) when starting from rest:

\[ s = ut + \frac{1}{2} a t^2 \]

Since \( u = 0 \):

\[ s = 0 + \frac{1}{2} a t^2 = \frac{1}{2} \times 2 \, \text{m/s}^2 \times (20 \, \text{s})^2 \]

Calculating this:

\[ s = \frac{1}{2} \times 2 \times 400 = 400 \, \text{m} \]

### Final Answer
The distance covered by the car is \( 400 \, \text{meters} \).

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