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, we need to analyze the motion of the particle given by the equation:

\[ y = 3(t - 2) + 5(t - 2)^2 \]

### Step 1: Differentiate the displacement equation to find the velocity.

The velocity \( v \) is given by the derivative of the displacement \( y \) with respect to time \( t \):

\[ v = \frac{dy}{dt} = \frac{d}{dt}[3(t - 2) + 5(t - 2)^2] \]

Using the chain rule and the power rule, we differentiate:

\[ v = 3 \cdot \frac{d}{dt}(t - 2) + 5 \cdot \frac{d}{dt}[(t - 2)^2] \]

Calculating each term:

1. The derivative of \( (t - 2) \) is \( 1 \).
2. The derivative of \( (t - 2)^2 \) is \( 2(t - 2) \).

Thus, we have:

\[ v = 3 \cdot 1 + 5 \cdot 2(t - 2) = 3 + 10(t - 2) \]

### Step 2: Find the velocity at \( t = 0 \).

Now we will substitute \( t = 0 \) into the velocity equation:

\[ v(0) = 3 + 10(0 - 2) = 3 - 20 = -17 \, \text{m/s} \]

### Step 3: Differentiate the velocity equation to find the acceleration.

Next, we differentiate the velocity equation to find the acceleration \( a \):

\[ a = \frac{dv}{dt} = \frac{d}{dt}[3 + 10(t - 2)] \]

Since the derivative of a constant is zero, we have:

\[ a = 10 \]

### Step 4: Check the position of the particle at \( t = 2 \).

Now we will substitute \( t = 2 \) into the original displacement equation to find the position:

\[ y(2) = 3(2 - 2) + 5(2 - 2)^2 = 3(0) + 5(0) = 0 \]

### Conclusion

From our calculations:
1. The initial velocity at \( t = 0 \) is \( -17 \, \text{m/s} \) (not \( 3 \, \text{m/s} \)).
2. The acceleration is \( 10 \, \text{m/s}^2 \) (not \( 5 \, \text{m/s}^2 \)).
3. The position at \( t = 2 \) is \( 0 \) (the particle is at the origin).

Thus, the correct statement is that the particle is at the origin at \( t = 2 \).