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Question: In the hare-tortoise race, the hare ran for \(2\;min\) at a speed of \(7.5Km/h\), slept for \(56\;mi...

In the hare-tortoise race, the hare ran for 2  min2\;min at a speed of 7.5Km/h7.5Km/h, slept for 56  min56\;min and again ran for 2  min2\;min at a speed of 7.5km/h7.5km/h. Find the average speed of the hare in the race.

Explanation

Solution

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=distancetimespeed=\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=distancetimespeed=\dfrac{distance}{time}
Average speed is the change in distance dd over a period of time tt, and is given as uavg=dtu_{avg}=\dfrac{d}{t}
Whereas, instantaneous speed is the small distance d  sd\;s covered during a small period of time d  td\;t. It is denoted using calculus and is given as uinst=dsdtu_{inst}=\dfrac{ds}{dt}
During the first run, the hare runs at a speed of v1=7.5Km/h×(10003600)=2.08m/sv_{1}=7.5Km/h\times \left(\dfrac{1000}{3600}\right)=2.08m/s over a time of T1=2  mins=120sT_{1}=2\;mins=120s
Then the distance covered, is given as d1=v1×T1=2.08×120=249.6md_{1}=v_{1}\times T_{1}=2.08\times 120=249.6m
During the sleep v2=0m/sv_{2}=0m/s for duration of T2=56  minT_{2}=56\;min and d2=0md_{2}=0m
Similarly, during the second run, the hare runs at a speed of v3=7.5Km/h×(10003600)=2.08m/sv_{3}=7.5Km/h\times \left(\dfrac{1000}{3600}\right)=2.08m/s over a time of T3=2  mins=120sT_{3}=2\;mins=120s
Then the distance covered, is given as d3=v3×T3=2.08×120=249.6md_{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=DTV=\dfrac{D}{T}
    V=249.6+249.660×60=499.23600=0.138m/s\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/s0.138m/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.