
In the adjoining figure is shown the time- distance \[\left( {t - s} \right)\] graph of a cyclist. Find out from the graph:
(i) Maximum speed of the cyclist.
(ii) Average speed in the whole journey.
Answer
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Hint: We know that motion is a change in the position of an object with time. In order to specify the position, we need to use a reference point and a set of axes. Graphical analysis is a convenient method of studying the motion of particles. It can be effectively applied to analyze the motion of a particle.
Complete step by step answer:
Consider a given graph; mark its distance as $OA$, $AB$ and $BC$ respectively,
In general, speed is a ratio of distance by time taken.
That is, ${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$
A cyclist travels a distance of \[20km\] that is $OA$ in $2$ hours then, speed is given by,
$s = \dfrac{d}{t}$
$ \Rightarrow s= \dfrac{{20}}{2}$
$\Rightarrow s= 10km{h^{ - 1}}$
And from $A$ to $B$, there is no distance travelled by the cyclist. Means, a cyclist is at rest for one hour.
Thus, speed is zero.
Next, form b to c cyclists travels a distance of $15km$ in $2$ hours. This is less than $20km$. Therefore,
Speed is given by,
$s = \dfrac{d}{t}$
$ \Rightarrow s=\dfrac{{15}}{2}$
$\Rightarrow s= 7.5km{h^{ - 1}}$
Thus, from these we can say, the maximum speed of the cyclist is $10km{h^{ - 1}}$. That is from $0$ to $a$.
Average speed is the ratio of total distance to the total time taken.
$ \Rightarrow {\text{average speed = }}\dfrac{{{\text{average distance}}}}{{{\text{total time}}}}$
Total distance travelled by the cyclist is $20 + 0 + 15 = 35km$
And total time taken=$5$ hours
Then, ${\text{average speed = }}\dfrac{{35}}{5} = 7km{h^{ - 1}}$
$\therefore$ The maximum speed of the cyclist is $10km{h^{ - 1}}$ and the average speed in the whole journey is $7km{h^{ - 1}}$.
Additional information:
For graphical representation, we require two coordinate axes. For usual representation we will consider the x-axis and y-axis. Distance-time graph is represented by plotting distance along the y-axis and time is along the x-axis. A straight line parallel to X-axis in the distance-time graph tells us an object is at rest.
Note:
The slope of distance-time gives the velocity of the particle.
Displacement of a particle is the shortest distance between the initial and final position of the body.
Distance is the total path length covered by the body.
Complete step by step answer:
Consider a given graph; mark its distance as $OA$, $AB$ and $BC$ respectively,
In general, speed is a ratio of distance by time taken.
That is, ${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$
A cyclist travels a distance of \[20km\] that is $OA$ in $2$ hours then, speed is given by,
$s = \dfrac{d}{t}$
$ \Rightarrow s= \dfrac{{20}}{2}$
$\Rightarrow s= 10km{h^{ - 1}}$
And from $A$ to $B$, there is no distance travelled by the cyclist. Means, a cyclist is at rest for one hour.
Thus, speed is zero.
Next, form b to c cyclists travels a distance of $15km$ in $2$ hours. This is less than $20km$. Therefore,
Speed is given by,
$s = \dfrac{d}{t}$
$ \Rightarrow s=\dfrac{{15}}{2}$
$\Rightarrow s= 7.5km{h^{ - 1}}$
Thus, from these we can say, the maximum speed of the cyclist is $10km{h^{ - 1}}$. That is from $0$ to $a$.
Average speed is the ratio of total distance to the total time taken.
$ \Rightarrow {\text{average speed = }}\dfrac{{{\text{average distance}}}}{{{\text{total time}}}}$
Total distance travelled by the cyclist is $20 + 0 + 15 = 35km$
And total time taken=$5$ hours
Then, ${\text{average speed = }}\dfrac{{35}}{5} = 7km{h^{ - 1}}$
$\therefore$ The maximum speed of the cyclist is $10km{h^{ - 1}}$ and the average speed in the whole journey is $7km{h^{ - 1}}$.
Additional information:
For graphical representation, we require two coordinate axes. For usual representation we will consider the x-axis and y-axis. Distance-time graph is represented by plotting distance along the y-axis and time is along the x-axis. A straight line parallel to X-axis in the distance-time graph tells us an object is at rest.
Note:
The slope of distance-time gives the velocity of the particle.
Displacement of a particle is the shortest distance between the initial and final position of the body.
Distance is the total path length covered by the body.
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