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Question: The position of a particle varies with time as \(x = a{t^2} - b{t^3}\) . The acceleration of particl...

The position of a particle varies with time as x=at2−bt3x = a{t^2} - b{t^3} . The acceleration of particle is zero at time T equal to
A) ab\dfrac{a}{b}
B) 2a3b\dfrac{{2a}}{{3b}}
C) a3b\dfrac{a}{{3b}}
D) Zero

Explanation

Solution

In this question, we can differentiate the given equation once then we can deduce the equation for velocity and if we differentiate the given equation twice then we can deduce the equation for acceleration.

Complete step by step solution:
In this question, we have an equation in distance and time.
He given equation is-
x=at2−bt3x = a{t^2} - b{t^3} ……………………………….. (i)
We know that velocity of a particle is given as-
v=dxdtv = \dfrac{{dx}}{{dt}}
So, differentiating the equation (i) with respect to t. We get-
dxdt=2at−3bt2\dfrac{{dx}}{{dt}} = 2at - 3b{t^2}.......................... (ii)
Now, we know that the acceleration of a particle is given as-
a = ddt(v)or a = ddt(dxdt)or a = d2xdt2 {\text{a = }}\dfrac{{\text{d}}}{{{\text{dt}}}}\left( {\text{v}} \right) {\text{or a = }}\dfrac{{\text{d}}}{{{\text{dt}}}}\left( {\dfrac{{{\text{dx}}}}{{{\text{dt}}}}} \right) {\text{or a = }}\dfrac{{{{\text{d}}^{\text{2}}}{\text{x}}}}{{{\text{d}}{{\text{t}}^{\text{2}}}}}
So, differentiating the equation (ii) with respect to t. We get-
d2xdt2=2a−6bt\dfrac{{{d^2}x}}{{d{t^2}}} = 2a - 6bt ……………………………...(iii)
If d2xdt2=0\dfrac{{{d^2}x}}{{d{t^2}}} = 0
Then from equation (iii), we get-
0=2a−6bt⇒2a=6bt⇒t=2a6b⇒t=a3b 0 = 2a - 6bt \Rightarrow 2a = 6bt \Rightarrow t = \dfrac{{2a}}{{6b}} \Rightarrow t = \dfrac{a}{{3b}}

Hence, option (C) is correct.

Additional information:
The equation of motion is generally given in the terms of distance(displacement) and time. So, with the help of the equation of motion, we can find the velocity and acceleration of the particle in that motion. Basically, the equation of motion gives all the basic information of the motion. We can also plot a graph on the basis of this equation. The displacement-time graph gives the velocity of the particle and the velocity-time graph gives the acceleration of the particle.

Note: In this question, we have to remember that if we differentiate the equation of distance with respect to time it gives the velocity of the particle. Similarly, we have to remember that if we differentiate the equation of velocity with respect to time it gives the acceleration of the particle.