The position of a particle as a function of time t, is given by x(t) = at + bt2 - ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

Question

The position of a particle as a function of time t, is given by x(t) = at + bt2 - ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be : (A) a + (b2/4c) (B) a + (b2/c) (C) a + (b2/2c) (D) a + (b2/3c)

Tardigrade Admin · Accepted Answer

x =at +bt2 -ct3 v = (dx/dt)=a+2bt-3ct2 a= (dv/dt) =2b-6ct =0 ⇒ t = (b/3c) v(at t= (b/3c)) = a +2b ((b/3c))-3c((b/3c)) =a +(b2/3c) .

Q. The position of a particle as a function of time t, is given by $x (t) = a t + b t^{2} - c t^{3}$
where $a, b$ and $c$ are constants. When the particle attains zero acceleration, then its velocity will be :

$a + \frac{b ^{2}}{4 c}$

$a + \frac{b ^{2}}{c}$

$a + \frac{b ^{2}}{2 c}$

$a + \frac{b ^{2}}{3 c}$

$x = a t + b t^{2} - c t^{3}$
$v = \frac{d x}{d t} = a + 2 b t - 3 c t^{2}$
$a = \frac{d v}{d t} = 2 b - 6 c t = 0 \Rightarrow t = \frac{b}{3 c}$
$v_{(a tt = \frac{b}{3 c})} = a + 2 b (\frac{b}{3 c}) - 3 c (\frac{b}{3 c})$
$= a + \frac{b ^{2}}{3 c} .$

Q. The position of a particle as a function of time t, is given by x(t)=at+bt2−ct3 where a,b and c are constants. When the particle attains zero acceleration, then its velocity will be :

a+4cb2​

a+cb2​

a+2cb2​

a+3cb2​