The displacement of a particle at time t is x, where x = t4-kt3 . If the velocity of the particle at time t = 2 is minimum, then

Question

The displacement of a particle at time t is x, where x = t4-kt3 . 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

Tardigrade Admin · Accepted Answer

Given, x=t4-k t3 ⇒ (d x/d t)=4 t3-3 k t2 ⇒ (d v/d t)=12 t2-6 k t At t=2, (d v/d t)= minimum =0 ∴ 12 × 22-6 k × 2=0 ⇒ 48-12 k=0 ⇒ k=4

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

k = 4

k = -4

k = 8

k = -8

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 $= 0$
$∴ 12 \times 2^{2} - 6 k \times 2 = 0$
$\Rightarrow 48 - 12 k = 0$
$\Rightarrow k = 4$

Q. The displacement of a particle at time t is x, where x=t4−kt3 . If the velocity of the particle at time t=2 is minimum, then