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
(a) 9 sec
(b) 5/3 sec
(c) 3/5 sec
(d) 2 sec

Answer 1

(a) 9 sec

$$\begin{gathered} s=45t+11t^2-t^3 \\ \Rightarrow \frac{ds}{dt}=45+22t-3t^2 \\ \text{According to the question,} \\ 3t^2-22t-45=0 \\ \Rightarrow 3t^2-27t+5t-45=0 \\ \Rightarrow 3t\left(t-9\right)+5\left(t-9\right)=0 \\ \Rightarrow \left(t-9\right)\left(3t+5\right)=0 \\ \Rightarrow \left(t-9\right)=0\mathrm{or}\left(3t+5\right)=0 \\ \text{As time can't be negative,} \\ t=9\sec \end{gathered}$$