Question

The position of a particle as a function of time t, is given by $x(t)= at + b{t}^{2} – c{t}^{3}$, where a, b and c are constants. When the particle attains zero acceleration, then it’s velocity will be?
A. $a + \dfrac {{b}^{2}}{4c}$
B. $a + \dfrac {{b}^{2}}{c}$
C. $a + \dfrac {{b}^{2}}{2c}$
D. $a + \dfrac {{b}^{2}}{3c}$

Answer 1

Hint: To solve this problem, first find the velocity of the particle. Differentiating the equation for position will give the velocity of the particle. Then, differentiate the equation for velocity to get the acceleration. Equate this equation to zero as acceleration is given as zero. Rearrange the equation and find the value of t. Substitute this value of t in the expression for velocity of the particle. This will give the velocity of the particle when acceleration is zero.
Formula used:
$v = \dfrac {dx}{dt}$
$a = \dfrac {dv}{dt}$

Complete answer:
The position of a particle is given by,
$x(t)= at + b{t}^{2} – c{t}^{3}$ ...(1)
Velocity of the particle is given by,
$v = \dfrac {dx}{dt}$
Differentiating equation. (1) w.r.t. time we get,
$v= a + 2bt – 3c{t}^{2}$ ...(2)
Acceleration is given by,
 $a = \dfrac {dv}{dt}$
Differentiating equation. (2) w.r.t. time we get,
$a = 2b – 6ct$ ...(3)
It is given that acceleration is zero. So, equation.(3) becomes
$0= 2b-6ct$
$\Rightarrow 6ct= 2b$
$\Rightarrow t = \dfrac {b}{3c}$ ...(4)
Substituting equation. (4) in equation. (3) we get,
$v= a + 2b \times \dfrac {b}{3c} – 3c \times {\dfrac {b}{3c}}^{2}$
$\Rightarrow v= a + \dfrac {2{b}^{2}}{3c}- \dfrac {{b}^{2}}{3c}$
$\Rightarrow v= a + \dfrac {{b}^{2}}{3c}$
Hence, when the particle attains zero acceleration, then it’s velocity will be $ a + \dfrac {{b}^{2}}{3c}$.

So, the correct answer is “Option D”.

Note:
To solve these types of problems, students should know the relation between velocity, positions and acceleration and should also know how to calculate velocity and acceleration from the position of the particle or vice-versa. They should also know basic mathematical tools like integration and differentiation.