Kinematics 5 Question 4

6. 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

(2019 Main, 09 April II)

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

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

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

(d) $a+\frac{b^{2}}{c}$

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Answer:

Correct Answer: 6. (c)

Solution:

  1. Position of particle is, $x(t)=a t+b t^{2}-c t^{3}$

So, its velocity is, $v=\frac{d x}{d t}=a+2 b t-3 c t^{2}$

and acceleration is, $a=\frac{d v}{d t}=2 b-6 c t$

Acceleration is zero, then $2 b-6 c t=0$

$\Rightarrow \quad t=\frac{2 b}{6 c}=\frac{b}{3 c}$

Substituting this ’ $t$ ’ in expression of velocity, we get

$v=a+2 b \Big(\frac{b}{3 c}\Big)-3 c \Big(\frac{b}{3 c}\Big)^{2}=a+\frac{2 b^{2}}{3 c}-\frac{b^{2}}{3 c}=a+\frac{b^{2}}{3 c}$



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