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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
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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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