Question

The expression for movement of a particle is given by $t = p{x^2} + qx$, where $t$ is the time taken by the particle and $x$ is the displacement of the particle. The acceleration of the particle at the origin is given by,
A. $\dfrac{{ - 2p}}{{{q^3}}}$
B. $\dfrac{{ - 2q}}{{{p^3}}}$
C. $\dfrac{{2p}}{{{q^3}}}$
D. $\dfrac{{2q}}{{{p^3}}}$

Answer 1

The velocity of a particle is given by, $v = \dfrac{{dx}}{{dt}}$ i.e. the rate of change of displacement with respect to time and the acceleration of a particle is given by, $a = \dfrac{{dv}}{{dt}}$ i.e. the rate of change of velocity with respect to time. In order to get to the acceleration of the particle first, the velocity is needed to be calculated as the rate of change of velocity is known as acceleration.

Complete Step by Step Answer:
Step 1.
The given expression for the motion of a particle is, $t = p{x^2} + qx$, where p and q are constants, $x$ is displacement and $t$ is time taken.

Step 2.
As the rate of change of displacement with respect to time is velocity, $v = \dfrac{{dx}}{{dt}}$.
Therefore, differentiating equation $t = p{x^2} + qx$ with respect to time, to get velocity,

$$\dfrac{{dt}}{{dt}} = \dfrac{{d\left( {p{x^2} + qx} \right)}}{{dt}} \\\ 1 = p\left( {2x \cdot \dfrac{{dx}}{{dt}}} \right) + q\left( {\dfrac{{dx}}{{dt}}} \right) \\\ 1 = 2px \cdot \left( {\dfrac{{dx}}{{dt}}} \right) + q\left( {\dfrac{{dx}}{{dt}}} \right) \\\$$

Step 3.
Since $\dfrac{{dx}}{{dt}} = v$ replace $\dfrac{{dx}}{{dt}}$ in the equation $1 = 2px \cdot \left( {\dfrac{{dx}}{{dt}}} \right) + q\left( {\dfrac{{dx}}{{dt}}} \right)$ with $v$, where $v$ is velocity and $\dfrac{{dx}}{{dt}}$ is the rate of change of displacement.
$1 = 2px \cdot v + qv$…………eq. (1)

Step 4.
As at the origin the x-coordinate is equal to $x = 0$ so put $x = 0$ in equation (1).
Equation (1) is, $1 = 2px \cdot v + qv$
Let’s put the value $x = 0$ in the above equation.

1 = 2p \cdot 0 \cdot v + qv \\\ 1 = 0 + q \cdot v \\\ 1 = q \cdot v \\\ v = \dfrac{1}{q} \\\ $$…………eq.(2) Step 5. Since, the rate of change of velocity is known as acceleration, $\dfrac{{dv}}{{dt}} = a$. Differentiating equation (1) with respect to time, in order to get acceleration. Therefore,

\dfrac{{d\left( 1 \right)}}{{dt}} = \dfrac{{d\left( {2px \cdot v + qv} \right)}}{{dt}} \\
0 = 2p \cdot \dfrac{{d\left( {x \cdot v} \right)}}{{dt}} + q \cdot \dfrac{{dv}}{{dt}} \\
0 = 2p \cdot \left( {v \cdot \dfrac{{dx}}{{dt}} + x \cdot \dfrac{{dv}}{{dt}}} \right) + q \cdot \dfrac{{dv}}{{dt}} \\

As we know that $\dfrac{{dx}}{{dt}} = v$ also$\dfrac{{dv}}{{dt}} = a$, so replace these values in above equation. $$0 = 2p \cdot \left( {v \cdot v + x \cdot a} \right) + q \cdot a$$…………eq.(3) Step 6. At origin the x-coordinate will be zero i.e.$x = 0$. Putting value $x = 0$ in equation (3), as we need to calculate the value of acceleration at origin. $$0 = 2p \cdot \left( {v \cdot v + x \cdot a} \right) + q \cdot a$$ Put, $x = 0$, $$0 = 2p \cdot \left( {v \cdot v + 0 \cdot a} \right) + q \cdot a$$ Let’s solve above equation and calculate the value of acceleration i.e.$a$,

0 = 2p \cdot \left( {v \cdot v + 0 \cdot a} \right) + q \cdot a \\
0 = 2p \cdot \left( {{v^2} + 0} \right) + q \cdot a \\
0 = 2p{v^2} + q \cdot a \\
a \cdot q = - 2p{v^2} \\
a = \dfrac{{ - 2p{v^2}}}{q} \\

Since, from equation (2) we know that$$v = \dfrac{1}{q}$$ Therefore, let’s replace the value of $$v$$ in equation (4) $$a = \dfrac{{ - 2p{v^2}}}{q}$$ Replacing the value$$v = \dfrac{1}{q}$$,

a = \left( {\dfrac{{ - 2p}}{q}} \right) \cdot \left( {\dfrac{1}{{{q^2}}}} \right) \\
a = \dfrac{{ - 2p}}{{{q^3}}} \\

**$\therefore$ The acceleration of the particle at the origin is $\dfrac{{-2p}}{{{q^3}}}$. The correct option is (A).** **Note:** While solving these type of questions students should remember to make proper equations as, after differentiating the given expression $t = p{x^2} + qx$ for the first time remember to put $x = 0$ in the relationship so as to get the relation between velocity ($v$) and $q$, because we need to replace the value of $v$ in the last step to get the desired result. Remember to use the chain rule as while calculating equation (3) we used chain rule because $x$ and $v$ both are variables.