Find the distance travelled by a body having velocity `v = 1 - t^(2)` from t = 0 to t = 2 sec .

Options

A. `3m`
B. `2m`
C. `1m`
D. `4m`

Explanation

To find the distance traveled by a body with a velocity function \( v(t) = 1 - t^2 \) from \( t = 0 \) to \( t = 2 \) seconds, we will follow these steps:

### Step 1: Set up the integral for distance
The distance \( s \) traveled by the body can be calculated using the integral of the velocity function over the given time interval. The formula for distance is:

\[ s = \int_{t_1}^{t_2} v(t) \, dt \]

In our case, \( t_1 = 0 \) and \( t_2 = 2 \). Thus, we have:

\[ s = \int_{0}^{2} (1 - t^2) \, dt \]

### Step 2: Evaluate the integral
Now, we will evaluate the integral:

\[ s = \int_{0}^{2} (1 - t^2) \, dt = \int_{0}^{2} 1 \, dt - \int_{0}^{2} t^2 \, dt \]

Calculating each part separately:

1. The integral of \( 1 \) from \( 0 \) to \( 2 \):

\[ \int_{0}^{2} 1 \, dt = [t]_{0}^{2} = 2 - 0 = 2 \]

2. The integral of \( t^2 \) from \( 0 \) to \( 2 \):

\[ \int_{0}^{2} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3} \]

### Step 3: Combine the results
Now we can combine the results of the two integrals:

\[ s = 2 - \frac{8}{3} \]

To combine these, we need a common denominator:

\[ s = \frac{6}{3} - \frac{8}{3} = \frac{6 - 8}{3} = \frac{-2}{3} \]

### Step 4: Interpret the result
Since distance cannot be negative, we need to check the velocity function over the interval \( [0, 2] \). The velocity \( v(t) = 1 - t^2 \) becomes zero when \( t = 1 \) (i.e., \( v(1) = 0 \)).

This means the body travels forward until \( t = 1 \) and then reverses direction from \( t = 1 \) to \( t = 2 \). Therefore, we need to calculate the distance traveled in both segments:

1. From \( t = 0 \) to \( t = 1 \):

\[ s_1 = \int_{0}^{1} (1 - t^2) \, dt \]

Calculating this:

\[ s_1 = \int_{0}^{1} (1 - t^2) \, dt = \left[ t - \frac{t^3}{3} \right]_{0}^{1} = \left[ 1 - \frac{1}{3} \right] = \frac{2}{3} \]

2. From \( t = 1 \) to \( t = 2 \):

\[ s_2 = \int_{1}^{2} (1 - t^2) \, dt = \left[ t - \frac{t^3}{3} \right]_{1}^{2} \]

Calculating this:

\[ s_2 = \left[ 2 - \frac{8}{3} \right] - \left[ 1 - \frac{1}{3} \right] = \left[ 2 - \frac{8}{3} \right] - \left[ 1 - \frac{1}{3} \right] \]

Calculating \( s_2 \):

\[ s_2 = \left[ 2 - \frac{8}{3} \right] - \left[ \frac{3}{3} - \frac{1}{3} \right] = \left[ \frac{6}{3} - \frac{8}{3} \right] - \left[ \frac{2}{3} \right] = \left[ -\frac{2}{3} \right] - \left[ \frac{2}{3} \right] = -\frac{4}{3} \]

### Final Distance Calculation
Thus, the total distance traveled is:

\[ \text{Total Distance} = s_1 + |s_2| = \frac{2}{3} + \frac{4}{3} = 2 \text{ meters} \]

### Conclusion
The distance traveled by the body from \( t = 0 \) to \( t = 2 \) seconds is **2 meters**.

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