Question

A particle moves in a straight line and its speed depends on time as $v = \left| {2t - 3} \right|$. Find the displacement of the particle in $5s$.

Answer 1

Recall that the displacement of a particle is defined as the shortest distance that is travelled by the particle between two points. The displacement of the particle depends on the initial and final position, and it is not necessary that the initial position should be at origin. Displacement of a particle is denoted by the symbol ‘s’ and can only be measured along a straight path. Also the displacement of a particle is always indicated by an arrow, where the arrow points in the direction of motion of the particle.

Complete step by step solution:
Given that the speed depends as $v = \left| {2t - 3} \right|$
Therefore, it can be written that
\Rightarrow v = \left\\{ {2t - 3,t \geqslant \dfrac{3}{2}} \right\\}
Or v = \left\\{ { - 2t + 3,t < \dfrac{3}{2}} \right\\}
Now displacement is the rate of change of velocity with respect to time. So it can be written that
$\Rightarrow s = \int_0^5 {vdt}$
$\Rightarrow s = \int\limits_0^{\dfrac{3}{2}} {vdt} + \int\limits_{\dfrac{3}{2}}^5 {vdt}$
$\Rightarrow \int\limits_0^{\dfrac{3}{2}} {( - 2t + 3)dt + \int\limits_{\dfrac{3}{2}}^5 {(2t - 3)dt} }$
$\Rightarrow [\dfrac{{ - 2{t^2}}}{2} + 3t]_0^{\dfrac{3}{2}} + [\dfrac{{2{t^2}}}{2} - 3t]_{\dfrac{3}{2}}^5$
Substituting the lower and upper limits, and solving we get
$\Rightarrow [ - \dfrac{9}{4} + \dfrac{9}{2}] + [25 - 15 - \dfrac{9}{4} + \dfrac{9}{2}]$
$\Rightarrow - \dfrac{9}{2} + \dfrac{9}{2} + 10 + \dfrac{9}{2}$
$\Rightarrow 10 + \dfrac{9}{2} = \dfrac{{20 + 9}}{2}$
$\Rightarrow \dfrac{{29}}{2}$

Therefore, the displacement of the particle in a time period of $5s$ is $\Rightarrow 14.5m$.

Note: It is important to note that the displacement of the particle has both magnitude and direction. So it is a vector quantity. Also the terms distance and displacement are not to be confused. They are both different terms. The displacement of the particle does not depend on the path and does not give any information about the path followed but the distance changes according to the path taken by the particle. Also the distance travelled by a particle can never be zero or negative. It is always positive. But the displacement of a particle can be positive, zero or negative.