The motion of a Particle moving along then y- axis is represented as `y = 3 ( t – 2) +5(t – 2 ) ^2` Identify the correct statement

Options

A. the initial (t =0) velocity of the particle is `3 ms ^(-1)`
B. the acceleration of the particle is `5 ms ^(-1)`
C. the particle is at the origin at t = 2 s
D. all of the above

Answer

To solve the problem step by step, we will analyze the motion of the particle given by the equation \( y = 3(t - 2) + 5(t - 2)^2 \).

### Step 1: Determine the Position of the Particle
The position of the particle at time \( t \) is given by the equation:
\[ y = 3(t - 2) + 5(t - 2)^2 \]

To check if the particle is at the origin at \( t = 2 \) seconds, we substitute \( t = 2 \) into the equation:
\[ y = 3(2 - 2) + 5(2 - 2)^2 \]
\[ y = 3(0) + 5(0)^2 \]
\[ y = 0 \]

So, at \( t = 2 \) seconds, the particle is at the origin \( (0, 0) \).

### Step 2: Calculate the Velocity of the Particle
Velocity is defined as the rate of change of position with respect to time, which is given by the derivative of \( y \) with respect to \( t \):
\[ v = \frac{dy}{dt} \]

Differentiating \( y \):
\[ v = \frac{d}{dt}[3(t - 2) + 5(t - 2)^2] \]
Using the chain rule:
\[ v = 3 \cdot 1 + 5 \cdot 2(t - 2) \cdot \frac{d}{dt}(t - 2) \]
\[ v = 3 + 10(t - 2) \]

Now, we will evaluate the velocity at \( t = 0 \):
\[ v(t = 0) = 3 + 10(0 - 2) \]
\[ v(t = 0) = 3 - 20 \]
\[ v(t = 0) = -17 \, \text{m/s} \]

### Step 3: Calculate the Acceleration of the Particle
Acceleration is defined as the rate of change of velocity with respect to time, which is the derivative of \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} \]

Differentiating \( v \):
\[ a = \frac{d}{dt}[3 + 10(t - 2)] \]
\[ a = 0 + 10 \cdot \frac{d}{dt}(t - 2) \]
\[ a = 10 \]

Thus, the acceleration is constant at \( 10 \, \text{m/s}^2 \).

### Conclusion
1. At \( t = 2 \) seconds, the particle is at the origin.
2. The velocity at \( t = 0 \) seconds is \( -17 \, \text{m/s} \).
3. The acceleration is \( 10 \, \text{m/s}^2 \).