91Ó°ÊÓ

When the displacement of an object during its motion changes by unequal amounts in equal time periods, then the object's velocity is variable. The average velocity of the object in a certain time period is calculated by taking the ratio of the change in position to the total time taken, i.e.,

$\begin{array}{c}\mathrm{Average}\mathrm{velocity}=\frac{\mathrm{Displacement}}{\mathrm{Total}\mathrm{time}\mathrm{taken}}\\ \stackrel{¯}{v}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}\end{array}$

The instantaneous velocity of an object is calculated by taking the limit of average velocity over the time period when time $\mathrm{Δ}t$becomes infinitesimally small.

The slope of the straight line formed by joining two points on the position-time graph gives the average velocity of the object between these points. And the slope of the tangent to the curve at any point gives the instantaneous velocity at that point.

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

Thus, the instantaneous velocity of the rabbit at t = 10.0 s is

From the graph, the position of the rabbit at t = 30.0 s is approximately equal to 16 m, and at t = 29.0 s, it is 15 m. The graph is a straight line between the small time period of 29 s and 30 s. The slope of this straight line can be approximated as the tangent on the graph at t = 30.0 s.

Therefore, the instantaneous velocity of the rabbit at t = 30.0 s is

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

From the graph, the position of the rabbit at t = 0 s is 0 m, and at t = 5.0 s, it is 2 m. The graph is a straight line between 0 s and 5.0 s. The slope of this straight line gives the value of average velocity.

Therefore, the average velocity of the rabbit between t = 0 s and t = 5.0 s is

$\begin{array}{c}\stackrel{¯}{v}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}\\ =\frac{\left(2-0\right)\mathrm{m}}{\left(5-0\right)\mathrm{s}}\\ =0.4{\mathrm{ms}}^{-1}.\end{array}$

From the graph, the position of the rabbit at t = 25.0 s is 9 m, and at t = 30.0 s, it is 16 m. The slope of the straight line joining the two points between 25.0 s and 30.0 s gives the average velocity.

Therefore, the average velocity of the rabbit between t = 25.0 s and t = 30.0 s is

$\begin{array}{c}\stackrel{¯}{v}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}\\ =\frac{\left(16-9\right)\mathrm{m}}{\left(30-25\right)\mathrm{s}}\\ =0.14{\mathrm{ms}}^{-1}.\end{array}$

From the graph, the position of the rabbit at t = 40.0 s is 20 m, and at t = 50.0 s, it is 10 m. The slope of the straight line joining the two points between 40.0 s and 50.0 s gives the average velocity.

Therefore, the average velocity of the rabbit between t = 40.0 s and t = 50.0 s is

$\begin{array}{c}\stackrel{¯}{v}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}\\ =\frac{\left(10-20\right)\mathrm{m}}{\left(50-40\right)\mathrm{s}}\\ =-1{\mathrm{ms}}^{-1}.\end{array}$

The value of average velocity is negative because the displacement of the rabbit is negative.