A particle moves along a straight line such that its displacement at any time t is given by `s = 3t^(3)+7t^(2)+14t + 5`. The acceleration of the particle at `t = 1s` is

Options

A. `18m//s^(2)`
B. `32m//s^(2)`
C. `29m//s^(2)`
D. `24m//s^(2)`

Answer

To find the acceleration of the particle at \( t = 1 \, \text{s} \), we start with the given displacement function:

\[ s(t) = 3t^3 + 7t^2 + 14t + 5 \]

### Step 1: Differentiate the displacement function to find velocity
The first step is to differentiate the displacement function \( s(t) \) with respect to time \( t \) to find the velocity \( v(t) \).

\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 5) \]

Using the power rule of differentiation:

\[ v(t) = 9t^2 + 14t + 14 \]

### Step 2: Differentiate the velocity function to find acceleration
Next, we differentiate the velocity function \( v(t) \) with respect to time \( t \) to find the acceleration \( a(t) \).

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2 + 14t + 14) \]

Again, using the power rule of differentiation:

\[ a(t) = 18t + 14 \]

### Step 3: Substitute \( t = 1 \) into the acceleration function
Now, we substitute \( t = 1 \, \text{s} \) into the acceleration function to find the acceleration at that moment.

\[ a(1) = 18(1) + 14 = 18 + 14 = 32 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle at \( t = 1 \, \text{s} \) is \( 32 \, \text{m/s}^2 \).

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