The displacement of a moving particle is given by, ` x=at^3 + bt^2 +ct + d`. The acceleration of particle at t=3 s would be

Options

A. 2(a+9b)
B. 2(9a+b)
C. 9(a+b)
D. 3(2a+b)

Answer

To find the acceleration of the particle at \( t = 3 \) seconds, we start with the given displacement equation:

\[ x = at^3 + bt^2 + ct + d \]

### Step 1: Find the Velocity
The velocity \( v \) of the particle is the first derivative of the displacement \( x \) with respect to time \( t \). Thus, we differentiate \( x \):

\[ v = \frac{dx}{dt} = \frac{d}{dt}(at^3 + bt^2 + ct + d) \]

Using the power rule for differentiation, we get:

\[ v = 3at^2 + 2bt + c \]

### Step 2: Find the Acceleration
The acceleration \( a \) of the particle is the derivative of the velocity \( v \) with respect to time \( t \):

\[ a = \frac{dv}{dt} = \frac{d}{dt}(3at^2 + 2bt + c) \]

Differentiating this expression, we have:

\[ a = 6at + 2b \]

### Step 3: Substitute \( t = 3 \) seconds
Now, we substitute \( t = 3 \) seconds into the acceleration equation:

\[ a = 6a(3) + 2b \]

This simplifies to:

\[ a = 18a + 2b \]

### Step 4: Final Expression
We can express the acceleration at \( t = 3 \) seconds as:

\[ a = 18a + 2b \]

This can be rewritten as:

\[ a = 2(9a + b) \]

Thus, the acceleration of the particle at \( t = 3 \) seconds is:

\[ a = 2(9a + b) \]