The position x of a particle varies with time (t) as

x = 3t²– 2t³

The acceleration of the particle will be zero at time

Question

The position x of a particle varies with time as x-at2-bt3- The acceleration of particle is zero time T equal to

Toppr Admin · Accepted Answer

X-at2-x2212-bt3-x21D2- Velocity -dxdt-2at-x2212-3bt2Acceleration-d2xdt2-ddt-2at-x2212-3b2-2a-x2212-6btor a-2a-x2212-6btMaking a-0-x21D2-2a-6bt-x21D2-t-dfrac2a6b-a3b-

The position $x$ of a particle varies with time as $x = a t^{2} - b t^{3}$ . The acceleration of particle is zero at time $T$ equal to

$x = a t^{2} - b t^{3}$
$\Rightarrow$ Velocity $= \frac{d x}{d t} = 2 a t - 3 b t^{2}$
Acceleration $= \frac{d^{2} x}{d t^{2}} = \frac{d}{d t} (2 a t - 3 b^{2}) = 2 a - 6 b t$
or $a = 2 a - 6 b t$
Making $a = 0$
$\Rightarrow 2 a = 6 b t$
$⇒t=dfrac2a6b=a3b$ .

The position x of a particle varies with time as x=at2−bt3. The acceleration of particle is zero at time T equal to

x=at2−bt3⇒ Velocity =dxdt=2at−3bt2Acceleration=d2xdt2=ddt(2at−3b2)=2a−6btor a=2a−6btMaking a=0⇒2a=6bt⇒t=dfrac2a6b=a3b.

The position $x$ of a particle varies with time as $x = a t^{2} - b t^{3}$ . The acceleration of particle is zero at time $T$ equal to

$x = a t^{2} - b t^{3}$
$\Rightarrow$ Velocity $= \frac{d x}{d t} = 2 a t - 3 b t^{2}$
Acceleration $= \frac{d^{2} x}{d t^{2}} = \frac{d}{d t} (2 a t - 3 b^{2}) = 2 a - 6 b t$
or $a = 2 a - 6 b t$
Making $a = 0$
$\Rightarrow 2 a = 6 b t$
$⇒t=dfrac2a6b=a3b$ .