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