1.15 Rectilinear motion  (Page 5/6)

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The representation of the variation of velocity with time, however, needs to be consistent with physical interpretation of motion. For example, we can not think of velocity - time plot, which is a vertical line parallel to the axis of velocity (Figure ii). Such plot is inconsistent as this would mean infinite numbers of values, against the reality of one velocity at a given instant. Similarly, the velocity-time plot should not be intersected by a vertical line twice as it would mean that the particle has more than one velocity at a given time (Figure i).

Area under velocity – time plot

The area under the velocity – time plot is equal to displacement. The displacement in the small time period “dt” is given by :

$\begin{array}{l}đx=vđt\end{array}$

Integrating on both sides between time intervals ${t}_{1}\phantom{\rule{5pt}{0ex}}\mathrm{and}\phantom{\rule{5pt}{0ex}}{t}_{2},$

$\begin{array}{l}\Delta x={\int }_{{t}_{1}}^{{t}_{2}}vđt\end{array}$

The right hand side integral graphically represents an area on a plot drawn between two variables : velocity (v) and time (t). The area is bounded by (i) v-t curve (ii) two time ordinates ${t}_{1}$ and ${t}_{2}$ and (iii) time (t) axis as shown by the shaded region on the plot. Thus, the area under v-t plot bounded by the ordinates give the magnitude of displacement (Δx) in the given time interval.

When v-t curve consists of negative values of velocity, then the curve extends into fourth quadrant i.e. below time axis. In such cases, it is sometimes easier to evaluate area above and below time axis separately. The area above time axis represents positive displacement, whereas area under time axis represents negative displacement. Finally, areas are added with proper sign to obtain the net displacement during the motion.

To illustrate the working of the process for determining displacement, let us consider the rectilinear motion of a particle represented by the plot shown.

Here,

$\begin{array}{l}\mathrm{Area of triangle OAB =}\frac{1}{2}x4x4=\mathrm{8 m}\\ \mathrm{Area of trapezium BCDE =}-\frac{1}{2}x2x\left(2+4\right)=\mathrm{-6 m}\\ \mathrm{Area of triangle EFG =}\frac{1}{2}x1x2=\mathrm{1 m}\\ ⇒\mathrm{Net area =}8-6+1=\mathrm{3 m}\end{array}$

A switch from positive to negative value of velocity and vice-versa is associated with change of direction of motion. It means that every intersection of the v – t curve with time axis represents a reversal of direction. Note that it is not the change of slope (from postive to negative and and vive versa) like on the position time that indicates a change in the direction of motion; but the intersection of time axis, which indicates change of direction of velocity. In the motion described in figure above, particle undergoes reversal of direction at two occasions at B and E.

It is also clear from the above example that displacement is given by the net area (considering appropriate positive and negative sign), while distance covered during the motion in the time interval is given by the cumulative area without considering the sign. In the above example, distance covered is :

s = Area of triangle OAB + Area of trapezium BCDE + Area of triangle EFG

$\begin{array}{l}⇒\mathrm{s =}8+6+1=\mathrm{15 m}\end{array}$

A person walks with a velocity given by |t – 2| along a straight line. Find the distance and displacement for the motion in the first 4 seconds. What is the average velocity in this period?

Here, the velocity is equal to the modulus of a function in time. It means that velocity is always positive. An inspection of the function reveals that velocity linearly decreases for the first 2 second from 2 m/s to zero. It, then, increases from zero to 2 m/s in the next 2 seconds. In order to obtain distance and displacement, we draw the Velocity – time plot plot as shown.

The area under the plot gives displacement. In this case, however, there is no negative displacement involved. As such, distance and displacement are equal.

$\begin{array}{l}\text{Displacement = ΔOAB + ΔBCD}\\ \mathrm{Displacement =}\frac{1}{2}x\mathrm{OB}x\mathrm{OA}+\frac{1}{2}x\mathrm{BD}x\mathrm{CD}\\ \mathrm{Displacement =}\frac{1}{2}x2x2+\frac{1}{2}x2x2=\mathrm{4 m}\end{array}$

and the average velocity is given by :

$\begin{array}{l}{v}_{\mathrm{avg}}=\frac{\mathrm{Displacement}}{\Delta t}=\frac{4}{4}=\mathrm{1 m/s}\end{array}$

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