The displacement of a particle is given by `y = a + bt + ct^2 - dt^4`. The initial velocity and acceleration are respectively.

Options

A. `b, -4d`
B. `-b, 2c`
C. `b,2c`
D. `2c, -4d`

Explanation

To find the initial velocity and acceleration of the particle whose displacement is given by the equation \( y = a + bt + ct^2 - dt^4 \), we will follow these steps:

### Step 1: Differentiate the displacement function to find the velocity function.
The displacement function is:
\[ y = a + bt + ct^2 - dt^4 \]

To find the velocity \( v(t) \), we differentiate \( y \) with respect to time \( t \):
\[ v(t) = \frac{dy}{dt} = \frac{d}{dt}(a + bt + ct^2 - dt^4) \]
Since \( a \) is a constant, its derivative is 0. The derivatives of the other terms are:
- \( \frac{d}{dt}(bt) = b \)
- \( \frac{d}{dt}(ct^2) = 2ct \)
- \( \frac{d}{dt}(-dt^4) = -4dt^3 \)

Thus, the velocity function is:
\[ v(t) = b + 2ct - 4dt^3 \]

### Step 2: Find the initial velocity.
The initial velocity \( v(0) \) is obtained by substituting \( t = 0 \) into the velocity function:
\[ v(0) = b + 2c(0) - 4d(0)^3 = b \]

### Step 3: Differentiate the velocity function to find the acceleration function.
Now, we differentiate the velocity function \( v(t) \) to find the acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(b + 2ct - 4dt^3) \]
Again, since \( b \) is a constant, its derivative is 0. The derivatives of the other terms are:
- \( \frac{d}{dt}(2ct) = 2c \)
- \( \frac{d}{dt}(-4dt^3) = -12dt^2 \)

Thus, the acceleration function is:
\[ a(t) = 2c - 12dt^2 \]

### Step 4: Find the initial acceleration.
The initial acceleration \( a(0) \) is obtained by substituting \( t = 0 \) into the acceleration function:
\[ a(0) = 2c - 12d(0)^2 = 2c \]

### Final Answer
The initial velocity and acceleration are:
- Initial Velocity: \( v(0) = b \)
- Initial Acceleration: \( a(0) = 2c \)

### Summary
Thus, the initial velocity and acceleration are \( b \) and \( 2c \) respectively.