Question

The distance $x$ covered by a particle varies with time $t$ as ${x^2} = 2{t^2} + 6t + 1$ . Its acceleration varies with $x$ as:
A. $x$
B. ${x^2}$
C. ${x^{ - 1}}$
D. ${x^{ - 3}}$
E. ${x^{ - 2}}$

Answer 1

You can start by differentiating the given equation with respect to time to get this equation $xv = 2t + 3$ . Then differentiate this equation with respect to time to get this equation $xa + {v^2} = 2$ . Then substitute the value of $v$ derived from the equation $xv = 2t + 3$ into the equation $xa + {v^2} = 2$ to reach the solution.

Complete step by step answer:
Before moving on to the mathematical calculations let’s first discuss what velocity and acceleration are alongside their equations.
Velocity – Velocity is the displacement of a body per unit time. Velocity in simpler terms represents how fast or slow a body is moving in reference to a fixed point. The equation for velocity is
$v = \dfrac{{dx}}{{dt}}$
Acceleration – Acceleration is the change in velocity of a body per unit time. Acceleration of a body is always due to an external force on the body. The equation for acceleration is
$a = \dfrac{{dv}}{{dt}}$
In the problem, we are given the following equation
${x^2} = 2{t^2} + 6t + 1$ (Equation 1)
Differentiating equation 1 with respect to time, we get
$\dfrac{{d{x^2}}}{{dt}} = \dfrac{{d(2{t^2} + 6t + 1)}}{{dt}}$
$\Rightarrow \dfrac{{d{x^2}}}{{dt}} \times \dfrac{{dx}}{{dx}} = \dfrac{{d(2{t^2} + 6t + 1)}}{{dt}}$
$\Rightarrow \dfrac{{d{x^2}}}{{dx}} \times \dfrac{{dx}}{{dt}} = \dfrac{{d(2{t^2} + 6t + 1)}}{{dt}}$
$\Rightarrow 2x\dfrac{{dx}}{{dt}} = 4t + 6$
$\because \dfrac{{dx}}{{dt}} = v$
$2xv = 4t + 6$
$xv = 2t + 3$ (Equation 2)
$v = \dfrac{{\left( {2t + 3} \right)}}{x}$
Squaring both sides, we get
${v^2} = \dfrac{{{{\left( {2t + 3} \right)}^2}}}{{{x^2}}}$
$\Rightarrow {v^2} = \dfrac{{4{t^2} + 12t + 9}}{{{x^2}}}$
$\Rightarrow {v^2} = \dfrac{{2\left( {2{t^2} + 6t + 1} \right) + 7}}{{{x^2}}}$
From equation 1, we know ${x^2} = 2{t^2} + 6t + 1$
$\Rightarrow {v^2} = \dfrac{{2{x^2} + 7}}{{{x^2}}}$ (Equation 4)
Differentiating equation 2 with respect to time, we get
$x\dfrac{{dv}}{{dt}} + v\dfrac{{dx}}{{dt}} = \dfrac{{d\left( {2t + 3} \right)}}{{dt}}$
$\Rightarrow xa + {v^2} = \dfrac{{d\left( {2t + 3} \right)}}{{dt}}$ $\left( {\because \dfrac{{dv}}{{dt}} = a} \right)$
$\Rightarrow xa + {v^2} = 2$ (Equation 5)
Substituting the value of ${v^2}$ from equation 4 in equation 5, we get
$\Rightarrow xa + \dfrac{{2{x^2} + 7}}{{{x^2}}} = 2$
$\Rightarrow {x^3}a + 7 = 0$
$\Rightarrow a = \dfrac{{ - 7}}{{{x^3}}}$
$\Rightarrow a = - 7{x^{ - 3}}$
Hence, the acceleration of the given object varies with ${x^{ - 3}}$ .
Hence, option D is the correct choice.

Note:
In the problem we took the long method, we could have also written the given equation as $x = \sqrt {2{t^2} + 6t + 1}$. Then differentiated this equation twice with respect to time, but the differentiation of this equation would involve complex mathematics, and the method used in the solution is long but involves easier calculations.