A car accelerates from rest with acceleration `1.2 m//s^(2).` A bus moves with constant speed of 12 mis in a parallel lane. How long does the car take from its start to meet the bus?

Options

A. 17 s
B. 8 s
C. 20 s
D. 12 s

Explanation

To solve the problem of how long it takes for the car to meet the bus, we can follow these steps:

### Step 1: Identify the given data
- The car starts from rest, so its initial velocity \( u = 0 \, \text{m/s} \).
- The acceleration of the car \( a = 1.2 \, \text{m/s}^2 \).
- The bus moves with a constant speed \( v = 12 \, \text{m/s} \).

### Step 2: Write the equations for distance traveled
1. **Distance traveled by the bus** in time \( t \):
\[ s_{\text{bus}} = v \cdot t = 12t \]

2. **Distance traveled by the car** in time \( t \) using the second equation of motion:
\[ s_{\text{car}} = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} (1.2) t^2 = 0.6 t^2 \]

### Step 3: Set the distances equal
Since the car meets the bus at the same distance, we can set the two equations equal to each other:
\[ s_{\text{car}} = s_{\text{bus}} \]
\[ 0.6 t^2 = 12t \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ 0.6 t^2 - 12t = 0 \]

### Step 5: Factor the equation
Factoring out \( t \):
\[ t(0.6t - 12) = 0 \]

### Step 6: Solve for \( t \)
This gives us two solutions:
1. \( t = 0 \) (the initial time when both start)
2. \( 0.6t - 12 = 0 \) leads to:
\[ 0.6t = 12 \implies t = \frac{12}{0.6} = 20 \, \text{s} \]

### Conclusion
The time taken by the car to meet the bus is \( t = 20 \, \text{s} \).

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