Question

The distance ‘s’ in meters travelled by a particle in ‘t’ seconds is given by s = \(\frac{2t^3}{3}-18t+\frac{5}{3}.\) The acceleration when the particle comes to rest is

• A. 12 m²/sec.
• B. 3 m²/sec.
• C. 18 m²/sec.
• D. 10 m²/sec.

Hint:

To find the acceleration of a particle, first find its velocity by differentiating the displacement function. Then, find when the particle comes to rest by setting the velocity to zero. Finally, find the acceleration by differentiating the velocity function. This method can be applied to various kinematic problems.

Answer 1

The Correct Option is A

The correct answer is (A) : 12 m²/sec.

Answer 2

The correct answer is: (A) 12 m²/sec.

The distance \( s \) in meters travelled by a particle in \( t \) seconds is given by:

\(s = \frac{2t^3}{3} - 18t + \frac{5}{3}\)

We are asked to find the acceleration when the particle comes to rest. 
Step 1: Find the velocity

The velocity \( v \) is the first derivative of the displacement function \( s(t) \) with respect to time \( t \). So, we differentiate \( s \) with respect to \( t \): \[ v = \frac{ds}{dt} = \frac{d}{dt}\left( \frac{2t^3}{3} - 18t + \frac{5}{3} \right) \] Differentiating each term: \[ v = 2t^2 - 18 \] Step 2: Find the time when the particle comes to rest

The particle comes to rest when the velocity \( v = 0 \). So, we set \( v = 0 \) and solve for \( t \): \[ 0 = 2t^2 - 18 \] Solving for \( t \): \[ 2t^2 = 18 \] \[ t^2 = 9 \] \[ t = 3 \text{ seconds} \quad (\text{since time cannot be negative}) \] Step 3: Find the acceleration

The acceleration \( a \) is the derivative of the velocity function \( v(t) \) with respect to time \( t \). So, we differentiate \( v \) with respect to \( t \): \[ a = \frac{dv}{dt} = \frac{d}{dt}(2t^2 - 18) \] Differentiating: \[ a = 4t \] Now, substitute \( t = 3 \) seconds into the acceleration formula: \[ a = 4(3) = 12 \, \text{m/s}^2 \] Therefore, the acceleration when the particle comes to rest is 12 m/s². 
Thus, the correct answer is (A) 12 m²/sec.