Question

A particle moves in a straight line with a velocity ${v_t} = \left| {t - 4} \right|\,{\text{m/s}}$ where $t$ is time in seconds. The distance covered by the particle in $8\,{\text{s}}$ is
A. $8\,{\text{m}}$
B. $16\,{\text{m}}$
C. $4\,{\text{m}}$
D. $2\,{\text{m}}$

Answer 1

Use the formula for the acceleration of the particle in terms of the velocity and the kinematic equation for the displacement of the particle. Determine the displacements of the particle for the first four seconds and then for the next four seconds.
Formula used:
The formula for the instantaneous acceleration $a$ of an object is
$a = \dfrac{{dv}}{{dt}}$ …… (1)
Here, $dv$ is the small change in the velocity and $dt$ is the small change in the time.
The kinematic expression for the displacement $s$ of an object is
$s = ut + \dfrac{1}{2}a{t^2}$ …… (2)
Here, $u$ is the initial velocity of the object, $a$ is the acceleration of the object and $t$ is the time.

Complete step by step answer:
he velocity ${v_t}$ of the particle is $\left| {t - 4} \right|\,{\text{m/s}}$.
${v_t} = \left| {t - 4} \right|\,{\text{m/s}}$ …… (3)
For the time $t < 4$, the velocity is $- \left( {t - 4} \right)$ and for time $t > 4$, the velocity is $t - 4$.
For time $t < 4$ (between $0\,{\text{s}}$ to $4\,{\text{s}}$), the velocity ${v_{t < 4}}$ is
${v_{t < 4}} = - t + 4$
Calculate the acceleration of the particle for the time $t < 4$.
Rewrite equation (1) for the acceleration ${a_{t < 4}}$ for the time $t < 4$.
${a_{t < 4}} = \dfrac{{d{v_{t < 4}}}}{{dt}}$
Substitute $- t + 4$ for ${v_{t < 4}}$ in the above equation.
${a_{t < 4}} = \dfrac{{d\left( { - t + 4} \right)}}{{dt}}$
$\Rightarrow {a_{t < 4}} = \dfrac{{d\left( { - t} \right)}}{{dt}} + \dfrac{{d\left( 4 \right)}}{{dt}}$
$\Rightarrow {a_{t < 4}} = \left( { - 1} \right) + \left( 0 \right)$
$\Rightarrow {a_{t < 4}} = - 1\,{\text{m/}}{{\text{s}}^2}$
Hence, the acceleration for time $t < 4$ is $- 1\,{\text{m/}}{{\text{s}}^2}$.
This acceleration is uniform. Hence, calculate the initial velocity at time $t = 0\,s$.
${v_{t < 4}} = - t + 4$
Calculate the displacement for the time $t = 0\,{\text{s}}$.
Substitute $0\,{\text{s}}$ for $t$ in equation (3).
${v_{0\,{\text{s}}}} = \left| {\left( {0\,{\text{s}}} \right) - 4} \right|\,{\text{m/s}}$
$\Rightarrow {v_{0\,{\text{s}}}} = 4\,{\text{m/s}}$
Hence, the initial velocity of the particle is $4\,{\text{m/s}}$.
Calculate the displacement for the time $t = 4\,{\text{s}}$.
Substitute $4\,{\text{s}}$ for $t$ in equation (3).
${v_{4\,{\text{s}}}} = \left| {\left( {4\,{\text{s}}} \right) - 4} \right|\,{\text{m/s}}$
$\Rightarrow {v_{4\,{\text{s}}}} = 0\,{\text{m/s}}$
Hence, the final velocity of the particle at $t = 4\,{\text{s}}$ is $0\,{\text{m/s}}$.
Calculate the displacement of the particle in the first 4 seconds.
Rewrite equation (2) for the displacement of the particle in the first four seconds.
${s_{t < 4}} = {v_{0\,{\text{s}}}}t + \dfrac{1}{2}{a_{t < 4}}{t^2}$
Substitute $4\,{\text{m/s}}$ for ${v_{0\,{\text{s}}}}$, $4\,{\text{s}}$ for $t$ and $- 1\,{\text{m/}}{{\text{s}}^2}$ for ${a_{t < 4}}$ in the above equation.
${s_{t < 4}} = \left( {4\,{\text{m/s}}} \right)\left( {4\,{\text{s}}} \right) + \dfrac{1}{2}\left( { - 1\,{\text{m/}}{{\text{s}}^2}} \right){\left( {4\,{\text{s}}} \right)^2}$
$\Rightarrow {s_{t < 4}} = \left( {16\,{\text{m}}} \right) - \left( {8\,{\text{m}}} \right)$
$\Rightarrow {s_{t < 4}} = 8\,{\text{m}}$
Hence, the displacement of the particle in the first four seconds is $8\,{\text{m}}$.
For time $t > 4$ (after $4\,{\text{s}}$), the velocity ${v_{t > 4}}$ is
${v_{t > 4}} = t - 4$
Calculate the acceleration of the particle for the time $t > 4$.
Rewrite equation (1) for the acceleration ${a_{t > 4}}$ for the time $t > 4$.
${a_{t > 4}} = \dfrac{{d{v_{t > 4}}}}{{dt}}$
Substitute $t - 4$ for ${v_{t > 4}}$ in the above equation.
${a_{t > 4}} = \dfrac{{d\left( {t - 4} \right)}}{{dt}}$
$\Rightarrow {a_{t > 4}} = \dfrac{{d\left( t \right)}}{{dt}} - \dfrac{{d\left( 4 \right)}}{{dt}}$
$\Rightarrow {a_{t > 4}} = \left( 1 \right) + \left( 0 \right)$
$\Rightarrow {a_{t > 4}} = 1\,{\text{m/}}{{\text{s}}^2}$
Hence, the acceleration for time $t > 4$ is $1\,{\text{m/}}{{\text{s}}^2}$.
The initial velocity in the interval between $4\,{\text{s}}$ to $8\,{\text{s}}$, ${v_{4\,{\text{s}}}}$ is the initial velocity of the particle.
Calculate the displacement of the particle in the interval between $4\,{\text{s}}$ to $8\,{\text{s}}$.
Rewrite equation (2) for the displacement of the particle in the first four seconds.
${s_{t > 4}} = {v_{4\,{\text{s}}}}t + \dfrac{1}{2}{a_{t > 4}}{t^2}$
Substitute $0\,{\text{m/s}}$ for ${v_{4\,{\text{s}}}}$, $4\,{\text{s}}$ for $t$ and $1\,{\text{m/}}{{\text{s}}^2}$ for ${a_{t > 4}}$ in the above equation.
${s_{t > 4}} = \left( {0\,{\text{m/s}}} \right)\left( {4\,{\text{s}}} \right) + \dfrac{1}{2}\left( {1\,{\text{m/}}{{\text{s}}^2}} \right){\left( {4\,{\text{s}}} \right)^2}$
$\Rightarrow {s_{t > 4}} = \left( {0\,{\text{m}}} \right) + \left( {8\,{\text{m}}} \right)$
$\Rightarrow {s_{t > 4}} = 8\,{\text{m}}$
Hence, the displacement of the particle in the interval between $4\,{\text{s}}$ to $8\,{\text{s}}$ is $8\,{\text{m}}$.
The total displacement $s$ of the particle is the sum of the displacements ${s_{t < 4}}$ and ${s_{t > 4}}$.
$s = {s_{t < 4}} + {s_{t > 4}}$
Substitute $8\,{\text{m}}$ for ${s_{t < 4}}$ and $8\,{\text{m}}$ for ${s_{t > 4}}$ in the above equation.
$s = \left( {8\,{\text{m}}} \right) + \left( {8\,{\text{m}}} \right)$
$\Rightarrow s = 16\,{\text{m}}$
Therefore, the total displacement of the particle is $16\,{\text{m}}$.

So, the correct answer is “Option B”.

Note:
One can also determine the displacement of the particle from the total area under the velocity-time curve of the given function.
This can also be seen from v=|t−4| that the velocity decreases continuously from 4m/s to zero in the first 4s and then increases from zero to 4m/s in the next 4s . Thus in the first 4s , the acceleration is $−1m/s^2$ and in the next 4s it is $1m/s^2$.