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

1. First, we determine when \(x = 4t^3 - 3t\) equals zero to verify statement (a):
   • Set \(4t^3 - 3t = 0\)
   • Factor the equation: \(t(4t^2 - 3) = 0\) 
   • Solutions: \(t = 0\) or \(4t^2 = 3 \Rightarrow t^2 = \frac{3}{4} \Rightarrow t = \pm0.866\)
2. Next, we analyze the velocity to evaluate statements (b), (c), and (d):
   • Velocity \(v = \frac{dx}{dt} = \frac{d}{dt}(4t^3 - 3t) = 12t^2 - 3\)
   • Set \(v = 0\) to find changing points: \(12t^2 - 3 = 0 \Rightarrow 12t^2 = 3 \Rightarrow t^2 = \frac{1}{4} \Rightarrow t = \pm0.5\)
   • Direction changes at \(t = \pm0.5\). Find corresponding \(x\):
     • For \(t = 0.5\): \(x = 4(0.5)^3 - 3(0.5) = 1 - 1.5 = -0.5\)
     • For \(t = -0.5\): \(x = 4(-0.5)^3 - 3(-0.5) = -0.5 + 1.5 = 1\)
3. Evaluate acceleration to assess statement (e):
   • Acceleration \(a = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 3) = 24t\)

Conclusion: The correct statements are (a), (c), and (e). Hence, the correct option is a, c, e.