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The Distance Moved by a Particle Travelling in Straight Line in T Seconds is Given by S = 45t + 11t2 − T3. the Time Taken by the Particle to Come to Rest is (A) 9 Sec (B) 5/3 Sec

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Question

The distance moved by a particle travelling in straight line in t seconds is given by s = 45t + 11t2 − t3. The time taken by the particle to come to rest is

Options

  • 9 sec

  •  5/3 sec

  • 3/5 sec

  • 2 sec

MCQ
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Solution

9 sec

\[s = 45t + 11 t^2 - t^3 \]

\[ \Rightarrow \frac{ds}{dt} = 45 + 22t - 3 t^2 \]

\[\text { According to the question,}\]

\[3 t^2 - 22t - 45 = 0\]

\[ \Rightarrow 3 t^2 - 27t + 5t - 45 = 0\]

\[ \Rightarrow 3t\left( t - 9 \right) + 5\left( t - 9 \right) = 0\]

\[ \Rightarrow \left( t - 9 \right)\left( 3t + 5 \right) = 0\]

\[ \Rightarrow \left( t - 9 \right) = 0 \text { or } \left( 3t + 5 \right) = 0\]

\[\text { As time can't be negative },\]

\[t = 9 \sec\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.4 [Page 25]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.4 | Q 13 | Page 25

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