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Question: A particle moving along a straight line has a velocity \(v\) \(m\) \({s^{ - 1}}\), when it cleared a...

A particle moving along a straight line has a velocity vv mm s−1{s^{ - 1}}, when it cleared a distance of xx mm. These two are connected by the relation v=49+xv = \sqrt {49 + x} . When its velocity is 1m1m s−1{s^{ - 1}}, its acceleration is
A. 2ms−22m{s^{ - 2}}
B. 7ms−27m{s^{ - 2}}
C. 1ms−21m{s^{ - 2}}
D. 0.5ms−20.5m{s^{ - 2}}

Explanation

Solution

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 vms−1vm{s^{ - 1}}
  2. The distance travelled by the particle D=xmD = xm
  3. There relation between the velocity of the particle and the distance travelled by it is given byv=49+xv = \sqrt {49 + x}
    We have to find the acceleration, now let us consider the following steps: -
    v=49+xv = \sqrt {49 + x} . . . (relation given in the question).
    On squaring both the sides, we have,
    v2=(49+x)2{v^2} = {\left( {\sqrt {49 + x} } \right)^2}
    v2=49+x{v^2} = 49 + x
    Now on differentiating with respect to time, we will get,
    2vdvdt=dxdt2v\dfrac{{dv}}{{dt}} = \dfrac{{dx}}{{dt}} . . . (1)
    (∵ddxxn=nxn−1)\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right) and
    (ddxconstan⁡t=0)\left( {\dfrac{d}{{dx}}cons\tan t = 0} \right)
    We know that
    The acceleration of the particle is a=dvdta = \dfrac{{dv}}{{dt}} . . . (2)
    and velocity of the particle is v=dxdtv = \dfrac{{dx}}{{dt}} . . . (3)
    Therefore, from equation (1), (2) and (3), we get
    2va=v2va = v
    ⇒a=v2v\Rightarrow a = \dfrac{v}{{2v}}
    ⇒a=12\Rightarrow a = \dfrac{1}{2}
    ⇒a=0.5m/s2\Rightarrow a = 0.5m/{s^2}
    Therefore, from the above explanation, the correct option is (D) 0.5m/s20.5m/{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:
v2−u2=2as{v^2} - {u^2} = 2as
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}