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

Answer

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

### Step 1: Find the expression for velocity
The velocity \( v \) of the particle is the rate of change of displacement with respect to time. This can be expressed mathematically as:
\[ v = \frac{dy}{dt} \]
Given the displacement function \( y = a + bt + ct^2 - dt^4 \), we differentiate it with respect to \( t \):
\[ v = \frac{d}{dt}(a + bt + ct^2 - dt^4) \]
Since \( a \) is a constant, its derivative is zero. Thus, we have:
\[ v = 0 + b + 2ct - 4dt^3 = b + 2ct - 4dt^3 \]

### Step 2: Calculate the initial velocity
To find the initial velocity, we evaluate the velocity expression at \( t = 0 \):
\[ v(0) = b + 2c(0) - 4d(0)^3 = b + 0 - 0 = b \]
Thus, the initial velocity \( v(0) \) is:
\[ \text{Initial Velocity} = b \]

### Step 3: Find the expression for acceleration
The acceleration \( a \) of the particle is the rate of change of velocity with respect to time. This can be expressed as:
\[ a = \frac{dv}{dt} \]
Using the velocity expression \( v = b + 2ct - 4dt^3 \), we differentiate it with respect to \( t \):
\[ a = \frac{d}{dt}(b + 2ct - 4dt^3) \]
Again, since \( b \) is a constant, its derivative is zero. Thus, we have:
\[ a = 0 + 2c - 12dt^2 = 2c - 12dt^2 \]

### Step 4: Calculate the initial acceleration
To find the initial acceleration, we evaluate the acceleration expression at \( t = 0 \):
\[ a(0) = 2c - 12d(0)^2 = 2c - 0 = 2c \]
Thus, the initial acceleration \( a(0) \) is:
\[ \text{Initial Acceleration} = 2c \]

### Final Answer
The initial velocity and acceleration are:
\[ \text{Initial Velocity} = b, \quad \text{Initial Acceleration} = 2c \]

### Summary
The initial velocity and acceleration of the particle are \( b \) and \( 2c \) respectively.

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