
In the hare-tortoise race, the hare ran for $2\;min$ at a speed of $7.5Km/h$, slept for $56\;min$ and again ran for $2\;min$ at a speed of $7.5km/h$. Find the average speed of the hare in the race.
Answer
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Hint: Speed is a scalar quantity which has only magnitude and not direction. We know that a speed is the change in the distance with respect to time. Clearly, speed depends on time. Then we can classify speed into average speed and instantaneous speed based on time. Here, we can find the time taken for the individual trips and then sum them to find the speed of the complete trip.
Formula used:
$speed=\dfrac{distance}{time}$
Complete step by step answer:
We know that speed is the change of distance with respect to time, and is given by $speed=\dfrac{distance}{time}$
Average speed is the change in distance $d$ over a period of time $t$, and is given as $u_{avg}=\dfrac{d}{t}$
Whereas, instantaneous speed is the small distance $d\;s$ covered during a small period of time $d\;t$. It is denoted using calculus and is given as $u_{inst}=\dfrac{ds}{dt}$
During the first run, the hare runs at a speed of $v_{1}=7.5Km/h\times \left(\dfrac{1000}{3600}\right)=2.08m/s$ over a time of $T_{1}=2\;mins=120s$
Then the distance covered, is given as $d_{1}=v_{1}\times T_{1}=2.08\times 120=249.6m$
During the sleep $v_{2}=0m/s$ for duration of $T_{2}=56\;min$ and $d_{2}=0m$
Similarly, during the second run, the hare runs at a speed of $v_{3}=7.5Km/h\times \left(\dfrac{1000}{3600}\right)=2.08m/s$ over a time of $T_{3}=2\;mins=120s$
Then the distance covered, is given as $d_{3}=v_{3}\times T_{3}=2.08\times 120=249.6m$
Then the average time is given as, the total distance covered by the total time taken, then we have, $V=\dfrac{D}{T}$
$\implies V=\dfrac{249.6+249.6}{60\times 60}=\dfrac{499.2}{3600}=0.138m/s$
Hence the average velocity of the hare is found to be $0.138m/s$
Note:
There is a small difference between velocity and speed. Velocity is a vector quantity which has both direction and magnitude, unlike speed which is a scalar quantity and can have only magnitude and not direction. This is one major difference between the both, and it is a physical quantity.
Formula used:
$speed=\dfrac{distance}{time}$
Complete step by step answer:
We know that speed is the change of distance with respect to time, and is given by $speed=\dfrac{distance}{time}$
Average speed is the change in distance $d$ over a period of time $t$, and is given as $u_{avg}=\dfrac{d}{t}$
Whereas, instantaneous speed is the small distance $d\;s$ covered during a small period of time $d\;t$. It is denoted using calculus and is given as $u_{inst}=\dfrac{ds}{dt}$
During the first run, the hare runs at a speed of $v_{1}=7.5Km/h\times \left(\dfrac{1000}{3600}\right)=2.08m/s$ over a time of $T_{1}=2\;mins=120s$
Then the distance covered, is given as $d_{1}=v_{1}\times T_{1}=2.08\times 120=249.6m$
During the sleep $v_{2}=0m/s$ for duration of $T_{2}=56\;min$ and $d_{2}=0m$
Similarly, during the second run, the hare runs at a speed of $v_{3}=7.5Km/h\times \left(\dfrac{1000}{3600}\right)=2.08m/s$ over a time of $T_{3}=2\;mins=120s$
Then the distance covered, is given as $d_{3}=v_{3}\times T_{3}=2.08\times 120=249.6m$
Then the average time is given as, the total distance covered by the total time taken, then we have, $V=\dfrac{D}{T}$
$\implies V=\dfrac{249.6+249.6}{60\times 60}=\dfrac{499.2}{3600}=0.138m/s$
Hence the average velocity of the hare is found to be $0.138m/s$
Note:
There is a small difference between velocity and speed. Velocity is a vector quantity which has both direction and magnitude, unlike speed which is a scalar quantity and can have only magnitude and not direction. This is one major difference between the both, and it is a physical quantity.
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