Question 4: Motion in 1D

Question

A particle starts moving from time $t=0$ and its coordinate is given as $x(t) = 4t^3 - 3t$
A. The particle returns to its original position (origin) 0.866 units later
B. The particle is 1 unit away from origin at its turning point
C. Acceleration of the particle is non-negative
D. The particle is 0.5 units away from origin at its turning point
E. Particle never turns back as acceleration is non-negative
Choose the correct answer from the options given below :

Accepted Answer

The original position at $\mathrm{t}=0$ is $\mathrm{x}(0)=4(0)^3-3(0)=0$.
To find when it returns to the origin :
$$ 4 t^3-3 t=0 $$
$$\Rightarrow $$ $t\left(4 t^2-3\right)=0$
$$ 4 t^2=3 \Rightarrow t^2=0.75 \Rightarrow t=\sqrt{0.75} \approx 0.866 \mathrm{~s} $$
Hence, the Statement A is correct.
A turning point occurs when velocity $\mathrm{v}(\mathrm{t})=0$.
$$ v(t)=\frac{d x}{d t}=12 t^2-3 $$
At turning point,
$$ \mathrm{v}(\mathrm{t})=0 \Rightarrow 12 \mathrm{t}^2=3 \Rightarrow \mathrm{t}^2=0.25 \Rightarrow \mathrm{t}=0.5 \mathrm{~s} $$
So, the position at turning point is,
$$ x(0.5)=4(0.5)^3-3(0.5)=4(0.125)-1.5=0.5-1.5=-1.0 \mathrm{~m} $$
The distance from the origin is $|-1.0|=1$ unit.
Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect because the particle does turn back at $\mathrm{t}=0.5 \mathrm{~s}$.
Acceleration $\mathrm{a}(\mathrm{t})$ is the derivative of velocity:
$$ a(t)=\frac{d v}{d t}=24 t $$
Since the particle starts moving from $\mathrm{t}=0$, for all $\mathrm{t} \geq 0$, the value of 24 t will always be greater than or equal to zero.
Hence, statement C is correct.
Therefore, statements $\mathrm{A}, \mathrm{B}$, and C and D are correct. The option 2 best represents the correct set.
Hence, the correct option is (B).

Answer

A, C, D Only

Answer

A, B, C Only

Answer

C, E Only

Answer

A, C Only

Motion in 1D

Choose the correct answer: