Sketch the v vs t graph for the object whose displacement as a function of time is given by Fig. 2-44.

Figure 2-44 Problems 57,58 and 59

In the position-time graph, the position is taken along the y-axis and time is taken along the x-axis. The slope of the straight line formed by joining two points on the position-time graph gives the average velocity of the object. The slope of the tangent to the curve at any point gives the instantaneous velocity at that point.

You can plot a velocity-time graph corresponding to the position-time graph by calculating the instantaneous velocity at every point.

The velocity of an object is constant in a time period in which the slope of the position-time graph is constant. The position-time graph is a straight line from t = 0 s to t = 20 s, approximately. Thus, the slope of the graph and the velocity of the object are constant between t = 0 s and t = 20 s.

In this region, the average velocity of the object is equal to the instantaneous velocity.

Thus, the average velocity of the object between t = 0 s and t = 10 s is

$\begin{array}{c}v=\frac{\Delta x}{\Delta t}\\ =\frac{\left(3-0\right)\mathrm{m}}{\left(10-0\right)\mathrm{s}}\\ =0.3{\mathrm{ms}}^{-1}.\end{array}$

The average velocity of the object between t = 0 s and t = 20 s is

$\begin{array}{c}\overline{v}=\frac{\Delta x}{\Delta t}\\ =\frac{\left(6-3\right)\mathrm{m}}{\left(20-10\right)\mathrm{s}}\\ =0.3{\mathrm{ms}}^{-1}.\end{array}$

Thus, the velocity is $0.3{\mathrm{ms}}^{-1}$between t = 0 and t = 20 s.

The slope of the graph is the steepest between 22 s and 29 s. Thus, the velocity of the object will be at its peak in this duration.

The graph can be approximated as a straight line between very small time periods. The position of the object at t = 25 s is approximately 9 m, and at t = 26 s, it is 9.5 m. Thus, the instantaneous velocity of the object at t = 25 s is

$\begin{array}{c}v=\frac{\left(9.5-9\right)\mathrm{m}}{\left(26-25\right)\mathrm{s}}\\ =0.5{\mathrm{ms}}^{-1}.\end{array}$

The position of the object at t = 30.0 s is approximately 16 m, and at t = 29.0 s, it is 15 m. Thus, the instantaneous velocity of the object at t = 30.0 s is

$\begin{array}{c}v=\frac{\left(16-15\right)\mathrm{m}}{\left(30-29\right)\mathrm{s}}\\ =1{\mathrm{ms}}^{-1}.\end{array}$

At t = 37 s, the velocity is zero as the tangent at this point is parallel to the time axis.

At t = 40 s, the instantaneous velocity of the object is

$\begin{array}{c}v=\frac{\left(19.5-20\right)\mathrm{m}}{\left(41-40\right)\mathrm{s}}\\ =-0.5{\mathrm{ms}}^{-1}.\end{array}$

At t = 45 s, the instantaneous velocity of the object can be approximated as

$\begin{array}{c}v=\frac{\left(14.5-15.5\right)\mathrm{m}}{\left(46-45\right)\mathrm{s}}\\ =-1{\mathrm{ms}}^{-1}.\end{array}$

At t = 50 s, the instantaneous velocity of the object can be approximated as

$\begin{array}{c}v=\frac{\left(10-10\right)\mathrm{m}}{\left(50-49\right)\mathrm{s}}\\ =0{\mathrm{ms}}^{-1}.\end{array}$

Using the calculated values of velocity, you can plot the graph between velocity and time as.