A car starts from rest, attain a velocity of `36 km h^(-1)` with an acceleration of `0.2 m s^(-2)`, travels 9 km with this uniform velocity and then comes to halt with a uniform deaceleration of 0.1 `m s^(-2)`. The total time of travel of the car is

Options

A. 1050 s
B. 1000 s
C. 950 s
D. 900 s

Explanation

To solve the problem step by step, we will break it down into three sections: acceleration, uniform motion, and deceleration.

### Step 1: Calculate Time to Accelerate to 36 km/h
1. **Convert the velocity from km/h to m/s**:
\[ v = 36 \text{ km/h} = \frac{36 \times 1000}{3600} = 10 \text{ m/s} \]

2. **Use the first equation of motion to find the time taken to reach this velocity**:
\[ v = u + at \]
Here, \( u = 0 \) (initial velocity), \( v = 10 \text{ m/s} \), and \( a = 0.2 \text{ m/s}^2 \).
\[ 10 = 0 + 0.2 \cdot t_1 \]
\[ t_1 = \frac{10}{0.2} = 50 \text{ seconds} \]

### Step 2: Calculate Time to Travel 9 km at Uniform Velocity
1. **Convert the distance from km to m**:
\[ d = 9 \text{ km} = 9000 \text{ m} \]

2. **Use the formula for time**:
\[ t_2 = \frac{d}{v} \]
\[ t_2 = \frac{9000 \text{ m}}{10 \text{ m/s}} = 900 \text{ seconds} \]

### Step 3: Calculate Time to Decelerate to a Stop
1. **Use the first equation of motion again for deceleration**:
\[ v = u + at \]
Here, \( v = 0 \) (final velocity), \( u = 10 \text{ m/s} \), and \( a = -0.1 \text{ m/s}^2 \) (deceleration).
\[ 0 = 10 - 0.1 \cdot t_3 \]
\[ 0.1 \cdot t_3 = 10 \]
\[ t_3 = \frac{10}{0.1} = 100 \text{ seconds} \]

### Step 4: Calculate Total Time of Travel
1. **Add the times from all three sections**:
\[ \text{Total time} = t_1 + t_2 + t_3 \]
\[ \text{Total time} = 50 + 900 + 100 = 1050 \text{ seconds} \]

### Final Answer
The total time of travel of the car is **1050 seconds**.

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