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Question: You move along \(+x\) direction through a distance of \(10\,m\) and then move back through a distanc...

You move along +x+x direction through a distance of 10m10\,m and then move back through a distance of 4m4\,m Repeat It four times during ten minutes, find the:
(A) total
(i)(i) Distance
(ii)(ii) Displacement and
(B) average
(i)(i) speed,
(ii)(ii) velocity.

Explanation

Solution

We come across scenarios where we need to determine which of the two or more things is going faster. The faster of the two, one can easily tell if they are going in the same direction on the same lane. But if their direction of motion is in the opposite direction, then deciding the fastest is difficult.

Formula used:
speed  =DistanceTimespeed\; = \dfrac{Distance}{Time}
Velocity  =DisplacementTimeVelocity\; = \dfrac{Displacement}{Time}
distance  =  speed  ×  time.distance\; = \;speed\; \times \;time.
Displacement can be calculated by measuring the final distance away from a point, and then subtracting the initial distance.
displacement = (covereddistance - startingdistance)×time{\text{displacement = (covered}}\,{\text{distance - starting}}\,{{distance) \times time}}
Displacement is key when determining velocity
Velocity = displacement/time

Complete step by step solution:
Let covered distance is 10m10\,m ,back through a distance 4m4\,m
We find the total distance, displacement,
Then we find Average speed and velocity,
Let take,
(i)(i) Distance=(10+4)×10time{\text{Distance}} = \left( {10 + 4} \right) \times 10time
On simplifying, We get,
\Rightarrow Distance=14×10{\text{Distance}} = 14 \times 10
Multiplying the above equation,
Here,
Distance is 140m140\,m
(ii)(ii) Displacement
Displacement=(104)×10Displacement = \left( {10 - 4} \right) \times 10
On simplifying, We get,
\Rightarrow Displacement=6×10Displacement = 6 \times 10
Multiplying the above equation,
Here, Displacement is 60m60\,\,m
(ii)averageSpeed=Distancetime(ii)\,average\,Speed = \dfrac{{Dis\tan ce}}{{time}}
Substituting the given value in above equation,
We get, Averagespeed=14010minAverage\,speed = \dfrac{{140}}{{10\min }}
We know that one minute is equal to sixty seconds,
Here, Averagespeed=14010×60Average\,speed = \dfrac{{140}}{{10 \times 60}}
Then, performing the dividing operation,
Cancel the specific numbers,
We get, Averagespeed=140.1m/sAverage\,speed = \dfrac{{14}}{{0.1}}m/s
(ii)(ii) Averagevelocity=DisplacementtimeAverage\,velocity = \dfrac{{Displacement}}{{time}}
Substituting the given value,
Averagevelocity=6010×60Average\,velocity = \dfrac{{60}}{{10 \times 60}}
On simplifying, We get,
\Rightarrow 0.1m/s0.1\,m/s
Thus, the Average velocity is 0.1m/s0.1\,m/s
Hence, The total Distance is 140m140\,m
The total Displacement is 60m60\,\,m
Average speed is 140.1m/s\dfrac{{14}}{{0.1}}m/s
Average velocity is 0.1m/s0.1\,m/s.

Note: For most of us, speed and velocity can be a little confusing. This speed gives us an idea of how quickly an object moves, whereas speed not only tells us its speed but also tells us the direction in which the body moves. We can define velocity as a function of the distance traveled, while velocity is a displacement function.