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On a distance time plot, average speed is equal to the slope of the the straight line, joining the end points of the motion, makes with the time axis. Note that average speed is equal to the slope of the chord (AB) and not that of the tangent to the curve.
$$\begin{array}{cc}{v}_{a}=\mathrm{tan}\theta =\frac{\Delta s}{\Delta t}& =\frac{\mathrm{BC}}{\mathrm{AC}}\end{array}$$
Problem : The object is moving with two different speeds ${v}_{1}$ and ${v}_{2}$ in two equal time intervals. Find the average speed.
Solution : The average speed is given by :
$$\begin{array}{c}{v}_{a}=\frac{\Delta s}{\Delta t}\end{array}$$
Let the duration of each time interval is t. Now, total distance is :
$$\begin{array}{c}\Delta s={v}_{1}t+{v}_{2}t\end{array}$$
The total time is :
$$\begin{array}{c}\Delta t=t+t=\mathrm{2t}\end{array}$$
$$\begin{array}{ccc}\Rightarrow & {v}_{a}=& \frac{{v}_{1}t+{v}_{2}t}{\mathrm{2t}}\end{array}$$
$$\begin{array}{ccc}\Rightarrow & {v}_{a}=& \frac{{v}_{1}+{v}_{2}}{2}\end{array}$$
The average speed is equal to the arithmetic mean of two speeds.
The object is moving with two different speeds ${v}_{1}$ and ${v}_{1}$ in two equal intervals of distances. Find the average speed.
$$\begin{array}{c}v=\frac{\Delta s}{\Delta t}\end{array}$$
Let "s" be distance covered in each time interval ${t}_{1}$ and ${t}_{2}$ . Now,
$$\begin{array}{c}\Delta s=s+s=\mathrm{2s}\end{array}$$
$$\begin{array}{ccc}\Delta t={t}_{1}+{t}_{2}=& \frac{s}{{v}_{1}}+& \frac{s}{{v}_{2}}\end{array}$$
$$\begin{array}{ccc}\Rightarrow & {v}_{a}=& \frac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}\end{array}$$
The average speed is equal to the harmonic mean of two velocities.
Instantaneous speed is also defined exactly like average speed i.e. it is equal to the ratio of total distance and time interval, but with one qualification that time interval is extremely (infinitesimally) small. This qualification of average speed has important bearing on the value and meaning of speed. The instantaneous speed is the speed at a particular instant of time and may have entirely different value than that of average speed. Mathematically,
Where $\Delta s$ is the distance traveled in time $\Delta t$ .
As $\Delta t$ tends to zero, the ratio defining speed becomes finite and equals to the first derivative of the distance. The speed at the moment ‘t’ is called the instantaneous speed at time ‘t’.
On the distance - time plot, the speed is equal to the slope of the tangent to the curve at the time instant ‘t’. Let A and B points on the plot corresponds to the time $t$ and $t+\Delta t$ during the motion. As $\Delta t$ approaches zero, the chord AB becomes the tangent AC at A. The slope of the tangent equals ds/dt, which is equal to the instantaneous speed at 't'.
$\begin{array}{ccccc}v=\mathrm{tan}\theta & =& \frac{\text{DC}}{\text{AC}}& =& \frac{\text{ds}}{\text{dt}}\end{array}$
The distance covered in the small time period dt is given by :
$\begin{array}{c}\mathrm{ds}=vdt\end{array}$
Integrating on both sides between time intervals ${t}_{1}$ and ${t}_{2}$ ,
The right hand side of the integral represents an area on a plot drawn between two variables, speed (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.
Alternatively, we can consider the integral as the sum of areas of small strip of rectangular regions (vxdt), each of which represents the distance covered (ds) in the small time interval (dt). As such, the area under speed - time plot gives the total distance covered in a given time interval.
The position – time plot is similar to distance – time plot in one dimensional motion. In this case, we can assess distance and speed from a position - time plot. Particularly, if the motion is unidirectional i.e. without any reversal of direction, then we can substitute distance by position variable. The example here illustrates this aspect of one-dimensional motion.
Problem The position – time plot of a particle’s motion is shown below.
Determine :
Solution
1. Average speed in the first 10 seconds
The particle covers a distance of 110 m. Thus, average speed in the first 10 seconds is :
$$\begin{array}{l}{v}_{a}=\frac{110}{10}=\mathrm{11\; m/s}\end{array}$$
2. Instantaneous speed at 5 second
We draw the tangent at t = 5 seconds as shown in the figure. The tangent coincides the curve between t = 5 s to 7 s. Now,
$$\begin{array}{l}v=\mathrm{slope\; of\; the\; tangent}=\frac{30}{2}=\mathrm{15\; m/s}\end{array}$$
3. Maximum speed
We observe that the slope of the tangent to the curve first increase to become constant between A and B. The slope of the tangent after point B decreases to become almost flat at the end of the motion. It means the maximum speed corresponds to the constant slope between A and B, which is is 15 m/s
4. Time instant(s) when average speed equals instantaneous speed
Average speed is equal to slope of chord between one point to another. On the other hand, speed is equal to slope of tangent at point on the plot. This means that slope of chord between two times is equal to tangent at the end point of motion. Now, as time is measured from the start of motion i.e. from origin, we draw a straight line from the origin, which is tangent to the curve. The only such tangent is shown in the figure below. Now, this straight line is the chord between origin and a point. This line is also tangent to the curve at that point. Thus, average speed equals instantaneous speeds at t = 9 s.
Problem : Two particles are moving with the same constant speed, but in opposite direction. Under what circumstance will the separation between two remains constant?
Solution : The condition of motion as stated in the question is possible, if particles are at diametrically opposite positions on a circular path. Two particles are always separated by the diameter of the circular path. See the figure below to evaluate the motion and separation between the particles.
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