Question

Motion of a particle is given by equation \(S=(3t^3+7t^2+14t+8)\ m\), The value of acceleration of the particle at \(t=1 \ sec\) is:

• A. \(10\  m/s^2\)
• B. \(32 \ m/s^2\)
• C. \(23\  m/s^2\)
• D. \(16\  m/s\)

Answer 1

The Correct Option is B

To find the acceleration of the particle at \( t = 1 \) second, we need to follow these steps:

1. First, determine the velocity of the particle as a function of time by differentiating the position function with respect to time.
2. Then find the acceleration by differentiating the velocity function with respect to time.
3. Finally, substitute \( t = 1 \) second into the acceleration function to get the desired value.

Step-by-Step Solution

Step 1: Differentiate the position function to get velocity.

The position of the particle as a function of time is given by:

S = 3t^3 + 7t^2 + 14t + 8.

Velocity, \( V(t) \), is the first derivative of position \( S \) with respect to time \( t \):

V(t) = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 8) = 9t^2 + 14t + 14.

Step 2: Differentiate the velocity function to get acceleration.

Acceleration, \( a(t) \), is the first derivative of velocity \( V(t) \) with respect to time \( t \):

a(t) = \frac{d}{dt}(9t^2 + 14t + 14) = 18t + 14.

Step 3: Substitute \( t = 1 \) second into the acceleration function.

a(1) = 18(1) + 14 = 18 + 14 = 32 \ m/s^2.

Conclusion:

The acceleration of the particle at \( t = 1 \) second is 32 \ m/s^2, confirming that the correct answer is indeed 32 \ m/s^2.