Question

From the displacement-time graph of a cyclist given in the figure, find the average velocity in the first four seconds.

(A) $0.5m{s^ - }^1$
(B) $5m{s^ - }^1$
(C) $2.5m{s^ - }^1$
(D) $0m{s^ - }^1$

Answer 1

Hint: Average velocity over a period of time is the amount by which the cyclist has displaced divided by the time taken to travel that distance. It can be zero if the cyclist comes back to his/her starting position at that time. The distance a cyclist covers before coming back to the starting position doesn’t matter.
Formula used:
We know that,
$Average\ velocity = \dfrac{{Total\ displacement}}{{Total\ time\ taken}}$
Or, $v = \dfrac{s}{t}$
Where, $v$ is the average velocity,
$s$ is the displacement,
And, $t$ is the time taken.

Complete step by step answer
When you observe this plot, point A is the starting point at $t = 0$.
At point C, we can observe that the cyclist has traveled $10m$ in $t = 3s$.
At point D, we can observe that the cyclist takes a rest. Thus, no change in displacement at $t = 5s$.
From this information, we can conclude that at$t = 4s$, the cyclist has displaced $10m$.
We know that,
$v = \dfrac{s}{t}$
Putting the numerical values from the plot, we get
$v = \dfrac{{10}}{4}$
$ \Rightarrow v = 2.5m{s^ - }^1$

Additional information There is a difference between distance and displacement. The cyclist can move $10m$ from the starting point and come back. In this case, displacement will be zero, but the distance traveled by the cyclist will be$2 \times 10 = 20m$. The same thing can be said about speed and velocity. Speed is the distance traveled over a given time while velocity is displacement over time. Distance and speed are scalars while displacement and velocity are vectors.

Note While reading the graph, take care of the units of displacement and time. It can be something other than meters and seconds. Always be careful about whether it is a distance-time graph or a displacement-time graph.