Question

A particle moves a distance x in time t according to equation ${x^2} = 1 + {t^2}$. The acceleration of particle is:
A) $\dfrac{1}{{{x^3}}}$
B) $\dfrac{1}{x} - \dfrac{1}{{{x^2}}}$
C) $- \dfrac{t}{{{x^2}}}$
D) $\dfrac{1}{x} - \dfrac{{{t^2}}}{{{x^3}}}$

Answer 1

The given equation is in terms of distance and time and we need to find acceleration. So, we can find the relationship of acceleration with distance and time which will be in the form of differentiation and then calculate the value of acceleration.
For differentiating, we will use:
$\dfrac{d}{{dt}}\left( {{x^n}} \right) = n{x^{n - 1}}\dfrac{{dx}}{{dt}}$
And the identity:
$\dfrac{d}{{dt}}(uv) = u\dfrac{{dv}}{{dt}} + v\dfrac{{du}}{{dt}}$

Complete step by step answer:
Relationship between acceleration and distance in terms of time can be given as:
Velocity = rate of change of distance.
$v = \dfrac{{dx}}{{dt}}$
Acceleration = rate of change of velocity.
$a = \dfrac{{dv}}{{dt}}$
Substituting value of v:
$a = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right) \\\ \Rightarrow a = \dfrac{{{d^2}x}}{{d{t^2}}}......(1) \\\$
🡪 Acceleration is double derivative of the distance x.
Now, the given equation is:
${x^2} = 1 + {t^2}$ ________ (2)
Taking square root both sides to get the equation for x
$x = {\left( {1 + {t^2}} \right)^{\dfrac{1}{2}}}$
To find the acceleration we need to find the double derivative of this equation.
Differentiating both sides with respect to t, we get:
$\dfrac{{dx}}{{dt}} = \dfrac{1}{2}{\left( {1 + {t^2}} \right)^{\dfrac{1}{2} - 1}}(2t)\left( {\because \dfrac{d}{{dt}}\left( {{x^n}} \right) = n{x^{n - 1}}\dfrac{{dx}}{{dt}}} \right) \\\ \implies \dfrac{{dx}}{{dt}} = t{\left( {1 + {t^2}} \right)^{\dfrac{{ - 1}}{2}}} \\\$
Differentiating both sides with respect to t again:
$\dfrac{{{d^2}x}}{{d{t^2}}} = t \times - \dfrac{1}{2}{\left( {1 + {t^2}} \right)^{ - \dfrac{3}{2}}}(2t) + {\left( {1 + {t^2}} \right)^{ - \dfrac{1}{2}}}\left( {\because \dfrac{d}{{dt}}(uv) = u\dfrac{{dv}}{{dt}} + v\dfrac{{du}}{{dt}}} \right) \\\ \implies \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{ - t}}{{{{\left( {1 + {t^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{1}{{{{\left( {1 + {t^2}} \right)}^{\dfrac{1}{2}}}}} \\\ \implies \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{ - t}}{{{{\left( {{x^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{1}{{{{\left( {{x^2}} \right)}^{\dfrac{1}{2}}}}}[from(2)] \\\ \implies \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{{ - t}}{{{x^3}}} + \dfrac{1}{x} \\\ or \\\ \implies \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{1}{x} - \dfrac{t}{{{x^3}}} \\\$
From (1):
$a = \dfrac{{{d^2}x}}{{d{t^2}}} \\\ \Rightarrow a = \dfrac{1}{x} - \dfrac{{{t^2}}}{{{x^3}}} \\\$
Therefore, for the given equation, the value of acceleration is $\dfrac{1}{x} - \dfrac{{{t^2}}}{{{x^3}}}$ and the correct option is (D)

So, the correct answer is “Option D”.

Note:
When any quantity is differentiated with respect to the time, it denotes the rate of change of that quantity.
We had to substitute the value of $\left( {1 + {t^2}} \right)$ from equation (2) to match the value with the given options, if no options were given, there would have been no need to replace, we would still have the right value.
Differentiation is just the opposite of integration. For acceleration’s relationship with distance we differentiate while for distance’s relationship with acceleration, we have to integrate.