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Standard XII

Mathematics

Global Maxima

The displacem...

Question

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

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k = - 4

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k = 8

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k = - 8

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Solution

The correct option is A $k = 4$
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 $t = 2, \frac{d v}{d t} = m i n i m u m = 0$
$∴ 12 \times 2^{2} - 6 k \times 2 = 0$
$\Rightarrow 48 - 12 k = 0$
$\Rightarrow k = 4$

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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

The correct option is A k=4Given, x=t4−kt3⇒dxdt=4t3−3kt2⇒dvdt=12t2−6ktAt t=2,dvdt=minimum=0∴12×22−6k×2=0⇒48−12k=0⇒k=4

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

The correct option is A $k = 4$
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 $t = 2, \frac{d v}{d t} = m i n i m u m = 0$
$∴ 12 \times 2^{2} - 6 k \times 2 = 0$
$\Rightarrow 48 - 12 k = 0$
$\Rightarrow k = 4$