The displacements along taxis and y-axis are represented by `x= 4t + 4t^(2) and y=5t`. The velocity of particle at t = 1s is equal to

Options

A. 4 units
B. 5 units
C. 13 units
D. 9 units

Explanation

To find the velocity of the particle at \( t = 1 \, \text{s} \), we need to follow these steps:

### Step 1: Determine the displacement equations
The displacement along the x-axis and y-axis are given as:
\[ x = 4t + 4t^2 \]
\[ y = 5t \]

### Step 2: Calculate the velocity in the x-direction
The velocity in the x-direction (\( V_x \)) is the derivative of the displacement in the x-direction with respect to time:
\[ V_x = \frac{dx}{dt} \]
Calculating the derivative:
\[ V_x = \frac{d}{dt}(4t + 4t^2) = 4 + 8t \]
Now, substituting \( t = 1 \):
\[ V_x = 4 + 8(1) = 4 + 8 = 12 \, \text{units} \]

### Step 3: Calculate the velocity in the y-direction
The velocity in the y-direction (\( V_y \)) is the derivative of the displacement in the y-direction with respect to time:
\[ V_y = \frac{dy}{dt} \]
Calculating the derivative:
\[ V_y = \frac{d}{dt}(5t) = 5 \]
Since this is a constant, \( V_y \) remains the same regardless of time:
\[ V_y = 5 \, \text{units} \]

### Step 4: Calculate the resultant velocity
To find the resultant velocity (\( V \)), we use the Pythagorean theorem since the x and y components are perpendicular:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ V = \sqrt{(12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, \text{units} \]

### Final Answer
The velocity of the particle at \( t = 1 \, \text{s} \) is \( 13 \, \text{units} \).

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