Question

The displacement of a particle at time t is $x$, where $x = t^{4}-kt^{3}$ . If the velocity of the particle at time $t = 2$ is minimum, then

• A. k = 4
• B. k = -4
• C. k = 8
• D. k = -8

Answer 1

Answer: (A)

k = 4

Answer 2

Given, $x=t^{4}-k t^{3}$
$\Rightarrow \frac{d x}{d t}=4 t^{3}-3 k t^{2}$
$\Rightarrow \frac{d v}{d t}=12 t^{2}-6 k t$
At $t=2, \frac{d v}{d t}=$ minimum $=0$
$\therefore 12 \times 2^{2}-6 k \times 2=0$
$\Rightarrow 48-12 k=0$
$\Rightarrow k=4$