Question

The displacement of a body varies with the time as $S=t^3+3t^2+2t-1$. If the velocity at $t=4sec$ is $2+12Km/s$, then find $k$

• A. $6$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is A $6$

given: variation of displacement of a body with time is $s=t^3+3t^2+2t-1$

differentiating both sides we get, $v$

$v=\frac{dv}{ds}=\frac{d(t^3+3t^2+2t-1)}{dt}=3t^2+6t+2$

${\frac{ds}{dt}}_{t=4}=3\times {\left(4\right)}^2+6\times 4+2=48+24+2=74$m/s

$\therefore v$ at $t=4$s $=74$m/s

given $v=2+12k$m/s

$\Rightarrow 2+12k=74\Rightarrow 12k=72\Rightarrow k=6$