CUET · MATHS · PYQ PAPER 2023
A car starts from a point \(P\) at time \(t=0\) seconds and stops at \(Q\).
The distance \(x\), in meters, covered by it in \(t\) seconds is given by \(x(t)=t^3\left(3-\frac{t^2}{5}\right)\).
The distance between \(P\) and \(Q\) is:
- A \(\frac{162}{3} m\)
- B \(\frac{162}{5} m\)
- C 162 m
- D \(\frac{162}{9} m\)
Answer & Solution
Correct Answer
(B) \(\frac{162}{5} m\)
Step-by-step Solution
Detailed explanation
\(v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(3t^3 - \frac{t^5}{5}\right) = 9t^2 - t^4\) \(v(t) = 0 \implies 9t^2 - t^4 = 0 \implies t^2(9 - t^2) = 0 \implies t = 3\) \(x(3) = 3^3\left(3-\frac{3^2}{5}\right)\) \(x(3) = 27\left(3-\frac{9}{5}\right) = 27\left(\frac{15-9}{5}\right)\)…
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