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

पर्याय

• 9 sec

•  5/3 sec
  
  

• 3/5 sec
  
  

• 2 sec

Answer 1

9 sec

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

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

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

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

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

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

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

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

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

\[t = 9 \sec\]