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In the one- dimensional motion of a particle, the relation between position $x$ and time $t$ s given by ${x^2} + 2x = t$ ( here $x > 0$ ). Choose the correct statement:
(A) The retardation of the particle is $\dfrac{{ - 1}}{{4{{\left( {x + 1} \right)}^3}}}$
(B) The uniform acceleration of the particle is $\dfrac{1}{{{{\left( {x + 1} \right)}^3}}}$
(C) The uniform velocity of the particle is $\dfrac{1}{{{{\left( {x + 1} \right)}^3}}}$
(D) The particle has a variable acceleration of $4t + 6$

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Last updated date: 29th May 2024
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Answer
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Hint:- The rate of change of displacement with time gives the velocity. Thus if we differentiate the displacement with respect to time will give velocity. And when we differentiate the velocity with respect to time will give acceleration.

Complete Step by step answer:
If a function is described as the position of a body as a function of time, then we can differentiate it with respect to time. The first derivative gives the velocity of the body. The second derivative gives the acceleration of the body.
The equation connecting the displacement $x$ and time $t$ is given as,
${x^2} + 2x = t$
Differentiating the above equation with respect to time,
$\
  \dfrac{d}{{dt}}\left( {{x^2} + 2x} \right) = \dfrac{d}{{dt}}\left( t \right) \\
  2x.\dfrac{{dx}}{{dt}} + 2.\dfrac{{dx}}{{dt}} = 1 \\
\ $
Differentiating the displacement with respect to time will give the velocity.
$\
  2x.v + 2.v = 1 \\
  \Rightarrow 2v\left( {x + 1} \right) = 1 \\
  \Rightarrow 2v = \dfrac{1}{{\left( {x + 1} \right)}} \\
  \Rightarrow v = \dfrac{1}{{2\left( {x + 1} \right)}} \\
\ $
Differentiating the velocity with respect to time will give the acceleration.
That is, $a = \dfrac{{dv}}{{dt}}$
Thus calculating, we get
$\
  a = \dfrac{d}{{dt}}\dfrac{1}{{2\left( {x + 1} \right)}} \\
   = \dfrac{1}{2}\dfrac{d}{{dt}}\dfrac{1}{{\left( {x + 1} \right)}} \\
   = \dfrac{1}{2}.\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}\dfrac{d}{{dt}}\left( {x + 1} \right) \\
   = \dfrac{1}{2}.\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{dx}}{{dt}} \\
   = \dfrac{1}{2}.\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}.v \\
\ $
Substitute the value for velocity in above expression.
$\
  a = \dfrac{1}{2}.\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{1}{{2\left( {x + 1} \right)}} \\
   = \dfrac{{ - 1}}{{4{{\left( {x + 1} \right)}^3}}} \\
\ $
The acceleration is calculated as $\dfrac{{ - 1}}{{4{{\left( {x + 1} \right)}^3}}}$.
The acceleration is a negative value. This implies the retardation or decrease in acceleration.
The retardation of the particle is $\dfrac{{ - 1}}{{4{{\left( {x + 1} \right)}^3}}}$

The answer is option A

Note: The negative value of the acceleration implies the retardation. If the acceleration is positive, the velocity will increase. When the acceleration is negative, the velocity will decrease.