The position x of a particle varies with timer as x=at2-bt3. The acceleration of the particle will be zero at time t equal to

Question

The position x of a particle varies with timer as x=at2-bt3. The acceleration of the particle will be zero at time t equal to (A)  (2a/3b)  (B)  (1/b)  (C)  (a/3b)  (D) 0

Tardigrade Admin · Accepted Answer

x=at2-bt3 Velocity v=(dx/dt)=2at-3bt2 Acceleration a=(d2x/dt2)=2a-6bt Substituting acceleration a=0 ⇒ 2a-6bt=0 t=(2a/6b)=(a/3b)

Q. The position x of a particle varies with timer as $x = a t^{2} - b t^{3} .$ The acceleration of the particle will be zero at time t equal to

$\frac{2 a}{3 b}$

$\frac{1}{b}$

$\frac{a}{3 b}$

0

$x = a t^{2} - b t^{3}$ Velocity $v = \frac{d x}{d t} = 2 a t - 3 b t^{2}$ Acceleration $a = \frac{d ^{2} x}{d t ^{2}} = 2 a - 6 b t$ Substituting acceleration $a = 0$ $\Rightarrow$ $2 a - 6 b t = 0$ $t = \frac{2 a}{6 b} = \frac{a}{3 b}$

Q. The position x of a particle varies with timer as x=at2−bt3. The acceleration of the particle will be zero at time t equal to

3b2a​

b1​

3ba​

0

x=at2−bt3 Velocity v=dtdx​=2at−3bt2 Acceleration a=dt2d2x​=2a−6bt Substituting acceleration a=0 ⇒ 2a−6bt=0 t=6b2a​=3ba​

Q. The position x of a particle varies with timer as $x = a t^{2} - b t^{3} .$ The acceleration of the particle will be zero at time t equal to

$\frac{2 a}{3 b}$

$\frac{1}{b}$

$\frac{a}{3 b}$

$x = a t^{2} - b t^{3}$ Velocity $v = \frac{d x}{d t} = 2 a t - 3 b t^{2}$ Acceleration $a = \frac{d ^{2} x}{d t ^{2}} = 2 a - 6 b t$ Substituting acceleration $a = 0$ $\Rightarrow$ $2 a - 6 b t = 0$ $t = \frac{2 a}{6 b} = \frac{a}{3 b}$