The position x of a particle varies with time, (t) as x = at2 - bt3. The acceleration will be zero at time t is equal to

Question

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

Tardigrade Admin · Accepted Answer

Distance (x) = at2- bt3. Therefore velocity (v) = (dx/dt) = (d/dt) (at2- bt3) =2at -3bt2 and acceleration = (dv/dt) = (d/dt) (2at- 3bt2) = 2a - 6bt = 0 or t = (2a/6b) = (a/3b).

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

$\frac{a}{3 b}$

zero

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

$\frac{a}{b}$

Distance $(x) = a t^{2} - b t^{3} .$
Therefore velocity $(v) = \frac{d x}{d t} = \frac{d}{d t} (a t^{2} - b t^{3})$
$= 2 a t - 3 b t^{2}$ and
acceleration $= \frac{d v}{d t} = \frac{d}{d t} (2 a t - 3 b t^{2}) = 2 a - 6 b t = 0$
or $t = \frac{2 a}{6 b} = \frac{a}{3 b} .$

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

3ba​