Question

A particle moves along a line by $s=\frac{1}{3}{t}^3-3t^2+8t+5$, it changes its direction when

• A. $t=1,t=2$
• B. $t=2,t=4$
• C. $t=0,t=4$
• D. $t=2,t=3$

Answer 1

Answer: (B)

$t=2,t=4$

A particle changes its direction, when, velocity is equal to zero

Given,

$s=\frac{1}{3}{t}^3-3t^2+8t+5$

Velocity,

$v=\frac{ds}{dt}=\frac{1}{3}\left(3t^2\right)-6t+8$

$\therefore v=t^2-6t+8$

when $v=0$ particle changes its direction,

$\Rightarrow t^2-6t+8=0$

$(t-2)(t-4)=0$

$\therefore t=2,4$