If the displacement of a particle is given by `s=2t^(3)-5t^(2)+4t-3`, then its acceleration is `14" ft/sec"^(2)` after to,e

Options

A. 3 sec
B. 4 sec
C. 1 sec
D. 2 sec

Answer

To solve the problem step by step, we will find the acceleration of the particle given its displacement function \( s(t) = 2t^3 - 5t^2 + 4t - 3 \).

### Step 1: Find the velocity of the particle
The velocity \( v(t) \) is the first derivative of the displacement \( s(t) \) with respect to time \( t \).

\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 5t^2 + 4t - 3) \]

Using the power rule for differentiation:

\[ v(t) = 6t^2 - 10t + 4 \]

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

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 10t + 4) \]

Again, using the power rule for differentiation:

\[ a(t) = 12t - 10 \]

### Step 3: Set the acceleration equal to 14 ft/sec²
We need to find the time \( t \) when the acceleration is 14 ft/sec².

\[ 12t - 10 = 14 \]

### Step 4: Solve for \( t \)
Rearranging the equation:

\[ 12t = 14 + 10 \]
\[ 12t = 24 \]
\[ t = \frac{24}{12} = 2 \]

### Conclusion
The acceleration of the particle is 14 ft/sec² after \( t = 2 \) seconds.

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