MHT CET · Maths · Differentiation

The displacement of a particle at the time \(\mathrm{t}\) is given by \(\quad s=\sqrt{1+t}\), then its acceleration 'a' is proportional to

A square of the velocity
B \(\sqrt[3]{S}\)
C \(\sqrt{S}\)
D cube of the velocity

Verified Solution

Answer & Solution

Correct Answer

(D) cube of the velocity

Step-by-step Solution

Detailed explanation

Given \(s=\sqrt{1+t}\)
Differentiating with respect to \(\mathrm{t}\), we get
\(\mathrm{v} =\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{1}{2 \sqrt{1+\mathrm{t}}} \)
\( \mathrm{a} =\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}}(1+\mathrm{t})^{\frac{-1}{2}}=\frac{1}{2}\left(-\frac{1}{2}\right)(1+\mathrm{t})^{\frac{-3}{2}} \)
\( =\left(\frac{-1}{4}\right)\left(\frac{1}{(1+\mathrm{t})^{\frac{3}{2}}}\right)=-2\left(\frac{1}{\left[(1+\mathrm{t})^{\frac{1}{2}}\right]^{3}}\right)\) \(=-2\left(\frac{1}{2 \sqrt{1+\mathrm{t}^{2}}}\right)^{3}=-2 \mathrm{v}^{3} \)

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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