Question

A particle moves along the x-axis in such a way that its position at time $t$ is given by $x(t) = \dfrac{{2 - t}}{{1 - t}}$ .What is the acceleration of the particle at time $t = 0$ .

Answer 1

To solve this type of question, one must know the basics of differentiation. We will simply apply differentiation as we know that the rate of change of displacement is velocity and then again for acceleration similarly, we will differentiate the obtained equation of velocity to get the required result.

Complete step by step answer:
As we know that the rate of change of displacement is velocity and according to the question, position is given so differentiating we will get the velocity.
$x(t) = \dfrac{{2 - t}}{{1 - t}}$
$\dfrac{d}{{dt}}x(t) = v(t)$
$v(t) = \dfrac{d}{{dt}}\left[ {\dfrac{{2 - t}}{{1 - t}}} \right]$
On applying u by v form of differentiation,

$$\Rightarrow v(t) = \dfrac{{\left( {1 - t} \right)\dfrac{d}{{dt}}\left[ {2 - t} \right] - \left( {2 - t} \right)\dfrac{d}{{dt}}\left[ {1 - t} \right]}}{{{{\left( {1 - t} \right)}^2}}} \\\ \Rightarrow v(t) = \dfrac{{\left( {1 - t} \right)\left( { - 1} \right) - \left( {2 - t} \right)\left( { - 1} \right)}}{{{{\left( {1 - t} \right)}^2}}} \\\$$

On further solving the equation,
$\Rightarrow v(t) = \dfrac{{t - 1 + 2 - t}}{{{{\left( {1 - t} \right)}^2}}}$
$\Rightarrow v(t) = \dfrac{1}{{{{\left( {1 - t} \right)}^2}}} = {\left( {1 - t} \right)^{ - 2}}$ -----(1)
Here, we got the equation of velocity and we know that rate of change of velocity is acceleration so differentiating the equation (1),
$\dfrac{d}{{dt}}v(t) = a(t)$
On further solving,
$\Rightarrow a(t) = - 2{\left( {1 - t} \right)^{ - 3}}( - 1)$
$\Rightarrow a(t) = \dfrac{2}{{{{\left( {1 - t} \right)}^3}}}$ -----(2)
As we have got the equation of acceleration and time and according to the question, we have to find the acceleration of the particle at $t = 0$ .
So, putting the value of $t = 0$ in equation (2).

$$a(t) = \dfrac{2}{{{{\left( {1 - t} \right)}^3}}} \\\ \Rightarrow a(0) = \dfrac{2}{{{{\left( {1 - 0} \right)}^3}}} \\\ \Rightarrow a(0) = \dfrac{2}{{{{\left( 1 \right)}^3}}} \\\ \therefore a(0) = 2m{\text{ }}{s^{ - 2}} \\\$$

So, the acceleration of the particle at $t = 0$ is $2m{\text{ }}{s^{ - 2}}$ .

Note: The velocity is the rate of change of displacement whereas the acceleration is the rate of change of velocity. Both the physical quantity is vector quantity because they both have magnitude as well as direction. Velocity may or may not be zero and in case of acceleration, as it is defined as a change in velocity or speed, acceleration cannot be zero. This explains why there should be some movement in order to accelerate.