MHT CET · Maths · Application of Derivatives

The distance ' \(\mathrm{s}\) ' in meters covered by a particle in \(\mathrm{t}\) seconds is given by \(s=2+27 t-t^3\). The particle will stop after
distance.

A 65 meters
B 80 meters
C 56 meters
D 60 meters

Verified Solution

Answer & Solution

Correct Answer

(C) 56 meters

Step-by-step Solution

Detailed explanation

\(\mathrm{s}=2+27 \mathrm{t}-\mathrm{t}^3\) and particle stops when its velocity is zero.
\(\therefore \frac{\mathrm{ds}}{\mathrm{dt}}=27-3 \mathrm{t}^2=0 \Rightarrow \mathrm{t}^2=9 \Rightarrow \mathrm{t}=3 \mathrm{sec} .\)
\(\therefore\) Distance covered in \(3 \mathrm{sec}\), is
\(\mathrm{S}_{(\mathrm{t}=3)}=2+27(3)-(3)^3=56 \mathrm{~m}\)

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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The distance ' \(\mathrm{s}\) ' in meters covered by a particle in \(\mathrm{t}\) seconds is given by \(s=2+27 t-t^3\). The particle will stop after distance.

Step-by-step Solution

The distance ' \(\mathrm{s}\) ' in meters covered by a particle in \(\mathrm{t}\) seconds is given by \(s=2+27 t-t^3\). The particle will stop after
distance.