A car starts from rest and accelerates at a constant rate in a straight line. In the first second, the car covers a distance of 2.0 meters. How much additional distance will the car cover during the second second of its motion ?

Options

A. 2.0 m
B. 6.0 m
C. 4.0 m
D. 8.0 m

Answer

To solve the problem of how much additional distance the car will cover during the second second of its motion, we can follow these steps:

### Step 1: Understand the Given Information
- The car starts from rest (initial velocity \( u = 0 \)).
- The distance covered in the first second (\( s_1 \)) is 2.0 meters.
- We need to find the distance covered during the second second of motion.

### Step 2: Use the Equation of Motion
We can use the second equation of motion to find the acceleration (\( a \)):
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values for the first second (\( t = 1 \) second):
\[ 2 = 0 \cdot 1 + \frac{1}{2} a (1^2) \]
This simplifies to:
\[ 2 = \frac{1}{2} a \]
Multiplying both sides by 2 gives:
\[ a = 4 \, \text{m/s}^2 \]

### Step 3: Find the Velocity After the First Second
Now, we can find the velocity of the car after the first second using the formula:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + 4 \cdot 1 = 4 \, \text{m/s} \]

### Step 4: Calculate the Distance Covered in the Second Second
To find the distance covered during the second second (\( s_2 \)), we can again use the equation of motion, but this time for the interval from \( t = 1 \) to \( t = 2 \). The initial velocity for this interval is \( v = 4 \, \text{m/s} \), and the time interval is 1 second:
\[ s_2 = vt + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s_2 = 4 \cdot 1 + \frac{1}{2} \cdot 4 \cdot (1^2) \]
This simplifies to:
\[ s_2 = 4 + 2 = 6 \, \text{meters} \]

### Conclusion
The additional distance the car will cover during the second second of its motion is **6 meters**.

---