Question

Velocity of a particle varies with time as v=t2- 8t + C, then the value of C so that
the particle never comes to rest. (Velocity is given in m/s and time is in sec.)
(a) >16
(b) <16
(0) > 4
(d) <4​

Answer 1

$\Large\boxed{\sf{\quad(b)\quad\!\textgreater\ 16\quad}}$

Given,

$\longrightarrow\sf{v(t)=t^2-8t+C}$

For $\sf{t=0,}$

$\longrightarrow\sf{v(0)=C}$

This implies $\sf{C\neq0.}$

Let $\sf{C\ \textgreater\ 0.}$ It means the velocity should always be positive.

$\longrightarrow\sf{t^2-8t+C\ \textgreater\ 0}$

$\longrightarrow\sf{t^2-8t\ \textgreater-C\quad\quad\dots(1)}$

For minimum $\sf{C,}$ we have to find the least (negatively maximum) value of $\sf{t^2-8t.}$

$\longrightarrow\sf{t^2-8t=t(t-8)}$

$\longrightarrow\sf{t^2-8t=(t-4+4)(t-4-4)}$

$\longrightarrow\sf{t^2-8t=(t-4)^2-4^2}$

$\longrightarrow\sf{t^2-8t=(t-4)^2-16}$

Since $\sf{x^2\geq0}$ for $\sf{x\in\mathbb{R},}$

$\longrightarrow\sf{t^2-8t\geq-16}$

Therefore, (1) becomes,

$\longrightarrow\sf{-16\ \textgreater-C}$

$\longrightarrow\sf{\underline{\underline{C\ \textgreater\ 16}}}$

If $\sf{C\ \textless\ 0,}$ we have to find the greatest (positively maximum) value of $\sf{t^2-8t}$ which goes to infinity. Hence $\sf{C\ \textless\ 0}$ is maximum.