TS EAMCET · Physics · Motion In One Dimension

Consider a particle moving along the positive direction of $X$ -axis. The velocity of the particle is given by $v = α \sqrt{x}$ ( $α$ is a positive constant). At time $t = 0$ , if the particle is located at $x = 0$ , the time dependence of the velocity and the acceleration of the particle are respectively.

A $\frac{α^{2}}{2} t$ and $\frac{α^{2}}{2}$
B $α^{2} t$ and $α^{2}$
C $\frac{α}{2} t$ and $\frac{α}{2}$
D $\frac{α^{2}}{4} t$ and $\frac{α^{2}}{4}$

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Answer & Solution

Correct Answer

(A) $\frac{α^{2}}{2} t$ and $\frac{α^{2}}{2}$

Step-by-step Solution

Detailed explanation

Given that, v=αx dxdt=αx ∴v=dxdt dxx=αdt ∫dxx=α∫dt ⇒ 2x=αt+c At t=0, x=0 ⇒ c=0 ∴ 2x=αt On Squaring, ⇒ 4x=α2t2 ⇒ x=α2t24 On differentiating,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Consider a particle moving along the positive direction of X-axis. The velocity of the particle is given by v=αx (α is a positive constant). At time t=0, if the particle is located at x=0, the time dependence of the velocity and the acceleration of the particle are respectively.

Step-by-step Solution