A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation `v = (5t^(2) - 3t + 2)m//s` . Now give the answer of following questions :
Velocity of particle when acceleration is zero is :

Options

A. `1 m//s`
B. `(35)/(20)m//s`
C. `(31)/(20)m//s`
D. `38m//s`

Explanation

To solve the problem, we need to find the velocity of the particle when its acceleration is zero. Let's go through the steps systematically.

### Step 1: Write down the velocity function
The velocity \( v \) of the particle as a function of time \( t \) is given by:
\[ v(t) = 5t^2 - 3t + 2 \quad \text{(in m/s)} \]

### Step 2: Find the acceleration function
Acceleration \( a \) is defined as the rate of change of velocity with respect to time. Therefore, we need to differentiate the velocity function with respect to \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(5t^2 - 3t + 2) \]
Calculating the derivative:
\[ a(t) = 10t - 3 \]

### Step 3: Set the acceleration to zero
To find the time when the acceleration is zero, we set the acceleration function equal to zero:
\[ 10t - 3 = 0 \]

### Step 4: Solve for \( t \)
Rearranging the equation gives:
\[ 10t = 3 \implies t = \frac{3}{10} \text{ seconds} \]

### Step 5: Substitute \( t \) back into the velocity function
Now that we have the time when acceleration is zero, we substitute \( t = \frac{3}{10} \) back into the velocity function to find the corresponding velocity:
\[ v\left(\frac{3}{10}\right) = 5\left(\frac{3}{10}\right)^2 - 3\left(\frac{3}{10}\right) + 2 \]
Calculating each term:
\[ = 5 \cdot \frac{9}{100} - 3 \cdot \frac{3}{10} + 2 \]
\[ = \frac{45}{100} - \frac{9}{10} + 2 \]
Converting \(-\frac{9}{10}\) to a fraction with a denominator of 100:
\[ -\frac{9}{10} = -\frac{90}{100} \]
Now substituting:
\[ = \frac{45}{100} - \frac{90}{100} + \frac{200}{100} \]
Combining the fractions:
\[ = \frac{45 - 90 + 200}{100} = \frac{155}{100} = \frac{31}{20} \text{ m/s} \]

### Final Answer
The velocity of the particle when the acceleration is zero is:
\[ v = \frac{31}{20} \text{ m/s} \]

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