Question

A particle moving along a straight line has a velocity $v$$m$${s^{ - 1}}$, when it cleared a distance of $x$$m$. These two are connected by the relation $v = \sqrt {49 + x} $. When its velocity is $1m$${s^{ - 1}}$, its acceleration is
A. $2m{s^{ - 2}}$
B. $7m{s^{ - 2}}$
C. $1m{s^{ - 2}}$
D. $0.5m{s^{ - 2}}$

Answer 1

Hint: Here we apply the concept of acceleration and the rate of change of velocity. So, we can differentiate velocity to find the acceleration.

Complete step by step answer:
We have the following data from the given question: -
1) A particle is moving with a velocity $vm{s^{ - 1}}$
2) The distance travelled by the particle $D = xm$
3) There relation between the velocity of the particle and the distance travelled by it is given by$v = \sqrt {49 + x} $
We have to find the acceleration, now let us consider the following steps: -
$v = \sqrt {49 + x} $ . . . (relation given in the question).
On squaring both the sides, we have,
${v^2} = {\left( {\sqrt {49 + x} } \right)^2}$
${v^2} = 49 + x$
Now on differentiating with respect to time, we will get,
$2v\dfrac{{dv}}{{dt}} = \dfrac{{dx}}{{dt}}$ . . . (1)
$\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$ and
$\left( {\dfrac{d}{{dx}}cons\tan t = 0} \right)$
We know that
The acceleration of the particle is $a = \dfrac{{dv}}{{dt}}$ . . . (2)
and velocity of the particle is $v = \dfrac{{dx}}{{dt}}$ . . . (3)
Therefore, from equation (1), (2) and (3), we get
$2va = v$
$ \Rightarrow a = \dfrac{v}{{2v}}$
$ \Rightarrow a = \dfrac{1}{2}$
$ \Rightarrow a = 0.5m/{s^2}$
Therefore, from the above explanation, the correct option is (D) $0.5m/{s^2}$.

Note: Knowing the basics of differentiation is very important for such types of problems. This question could have been solved by using the laws of motion. But using differentiation makes it short. The acceleration of a moving particle travelling in a straight line varies with its displacement. Relation between acceleration, displacement and velocity, using the laws of motion, can be give as:
${v^2} - {u^2} = 2as$
$s = ut + \dfrac{1}{2}a{t^2}$