Question

For a particle moving in x-direction according to the relation \(x = 4t^3 - 3t\), consider the following statements:

[(a)] At \(t = 0.866\), \(x = 0\)
[(b)] Direction of velocity of particle remains same
[(c)] Direction of velocity of particle changes at \(x = -1\)
[(d)] Direction of velocity of particle changes at \(x = 0.5\,\text{m}\)
[(e)] Acceleration is non-negative
Correct statements are:

• A. a, c, e
• B. a, b, c
• C. a, b
• D. a, b, c, d

Hint:

Direction of motion changes when {velocity becomes zero}, not when position becomes zero — always check \(v(t)\).

Answer 1

The Correct Option is A

Concept:
Velocity is the first derivative of position with respect to time.
Acceleration is the second derivative of position with respect to time. 
Direction of velocity changes when velocity becomes zero. 
Step 1: Find Velocity
Given: \[ x = 4t^3 - 3t \] \[ v = \frac{dx}{dt} = 12t^2 - 3 \] 
Step 2: Check Statement (a)
Put \(x = 0\): \[ 4t^3 - 3t = 0 \Rightarrow t(4t^2 - 3) = 0 \] \[ t = 0,\quad t = \pm\sqrt{\frac{3}{4}} = \pm 0.866 \] Hence, at \(t = 0.866\), \(x = 0\). Statement (a) is correct
. Step 3: Check Direction Change of Velocity
Velocity becomes zero when: \[ 12t^2 - 3 = 0 \Rightarrow t^2 = \frac{1}{4} \Rightarrow t = \pm \frac{1}{2} \] At \(t = \frac{1}{2}\): \[ x = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{3}{2} = -1 \] Thus, direction of velocity changes at \(x = -1\). Statement (c) is correct
and (d) is incorrect
. Statement (b) is incorrect
since velocity changes sign. 
Step 4: Check Acceleration
\[ a = \frac{dv}{dt} = 24t \] Acceleration is zero at \(t=0\) and positive for \(t>0\). Thus, acceleration is non-negative
. Statement (e) is correct
. 
Final Conclusion:
Correct statements are \(\boxed{a,\,c,\,e}\).