Question

The equation of particle is $x={{t}^{3}}-9{{t}^{2}}+3t+1$ and $v=-24$, then a=
A. 3
B. 2
C. 1
D. 0

Answer 1

Hint:We know that velocity is the rate of change of displacement and acceleration is the rate of change of velocity. Here we have the function of displacement with time. To find acceleration we have to take the second derivative of the displacement function. We have the value of velocity so that we can find the time from which it is easy to find acceleration of the particle.

Formula used:
$v=\dfrac{dx}{dt}
Here, $v$ is the velocity of the particle, $dx$ is displacement and $dt$ is the small duration of time.

Complete step by step solution:
Displacement is the straight line distance covered by an object. The equation of motion of a particle is usually expressed in terms of distance covered by the particle in time. Velocity of the particle is the distance covered by a particle in unit time. Mathematically, it is the time derivative of distance. Acceleration is the time derivative of velocity or can also be expressed as the second derivative of distance with respect to time.

Given in the question, the equation of particle in terms of distance covered in t time as:
$x={{t}^{3}}-9{{t}^{2}}+3t+1$
Forgetting the equation for velocity we differentiate equation of motion with respect to time
$v=\dfrac{dx}{dt}=3{{t}^{2}}-18t+3$
Given velocity as: v= -24
That is,
$-24=3{{t}^{2}}-18t+3$
On solving this quadratic equation, we get time as:
Time, t=3s
Now for finding the equation of acceleration we differentiate the equation for velocity with time. Now we get acceleration as:
$a=6t-18$
Putting the value of time in this equation, we get acceleration of the particle as zero

Therefore, the correct answer is option D.

Note: Remember that velocity given this question is just to find the time. It is just a condition to find out time. By seeing velocity as a constant value don’t skip to the conclusion that acceleration will be zero since it may not be the same in all cases.