Question

The velocity of a particle moving on the x-axis is given by $v = {x^2} + x$ where $v\,in\,m/s$ and x is in m. Its acceleration in $m/{s^2}$ when passing through the point $x = 2m$.
A. 0
B. 5
C. 11
D. 30

Answer 1

Hint: Here a particle is moving on the x-axis with the given velocity which is expressed in the equation of distance and we have to find acceleration when it passes to a point that is at 2m distance. Start with the relation between the velocity, acceleration and distance and put all the values from the question in the relation.

Formula used:
The relation of the acceleration and change in velocity with distance is,
$a = v\dfrac{{dv}}{{dx}}$
Here, $a$ is the acceleration, $v$ is the velocity and $\dfrac{{dv}}{{dx}}$ is the change in velocity with distance.

Complete step by step solution:
First start with the basic information provided in the question:
Velocity of the particle moving on the x-axis,
$v = {x^2} + x$
Where v is velocity and x is the distance given.
Now we know that, acceleration is equal to the derivative of the velocity:
$\dfrac{{dv}}{{dx}} = 2x + 1$

We know that acceleration
$a = v\dfrac{{dv}}{{dx}} \\
\Rightarrow a = ({x^2} + x)\dfrac{{d({x^2} + x)}}{{dx}}$....(putting value of v in formula of acceleration)
By solving we get,
$a = ({x^2} + x)(2x + 1)$
Putting value of x from the question that is 2m
$a = ({2^2} + 2)(2 \times 2 + 1) = 30$
Therefore, $a = 30\,m/{s^2}$

Hence, the correct answer is option D.

Note: Be careful about the direction in which the particle is moving, here it is moving on the x-axis. Put all the values correctly in the formula of the acceleration otherwise the answer may differ. Value of distance was given in the question if it is not given that we cannot get the value of acceleration of the particle.