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A man runs $200m$ in first $25s$ and then turns back and runs $100m$ in next $15s$ towards the starting point. Find out his
A. Average velocity.
B. His average speed.

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Last updated date: 13th Jul 2024
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Answer
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Hint: Kinematics is the study of motion in a straight line direction. Straight line motion with continuous acceleration is the most basic accelerated motion. The velocity varies at the same rate during the motion in this case. This is a unique circumstance, but one that happens often in nature.

Complete step-by-step solution:
Let us know about average velocity. An object's average velocity is calculated by dividing its total displacement by the total time taken. To put it another way, it's the speed at which something travels from one place to another. A vector quantity is average velocity. Meters per second is the SI unit of measurement. When applicable, any distance unit per any time unit, such as miles per hour (mph) or kilometres per hour (km/h), can be used (kmph).
\[Average{\text{ }}velocity = \dfrac{{Total\,Displacement}}{{Total\,time}}\]
Where $x({t_i})$ is the position of the particle at time ${t_{i.}}$ .
The distance travelled divided by the time taken equals the average speed:
$Average\,speed = \dfrac{{Total\,dis\tan ce}}{{Total\,time}}$
Now let us come to the problem:
Given:
In first case,
Displacement $(S) = 200\,m$
  Time is given ${t_1} = \,25\,\sec $
In second case,
Displacement $(S) = \, - 100\,m$
Time is given ${t_2} = 15\,\sec $
As we know form the above discussion:
(a)
\[Average{\text{ }}velocity = \dfrac{{Total\,Displacement}}{{Total\,time}}\]
$Average\,velocity = \dfrac{{S + ( - S)}}{{{t_1} + {t_2}}}$
$Average\,velocity = \dfrac{{200 + ( - 100)}}{{25 + 15}}$
$Average\,velocity = 2.5\,m{s^{ - 1}}$
Now let us find the Average speed:
(b)
$Average\,speed = \dfrac{{Total\,dis\tan ce}}{{Total\,time}}$
$Average\,speed = \dfrac{{S + S}}{{{t_1} + {t_2}}}$
$Average\,speed = \dfrac{{200 + 100}}{{25 + 15}}$
$Average\,speed = 10\,m{s^{ - 1}}$

 Note:Let us get some idea about displacement and distance. The term "distance" refers to "how much ground an object has travelled" during its motion. Displacement is a vector quantity that describes "how far an object is out of place"; it is the object's total shift in location.