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When two bodies are in motion in the very same direction, the difference in between their speed gives the relative rate of the relocating objects. Consider 2 objects relocating via speeds of x and also y in the exact same direction. Then the family member rate is x-y.

When 2 bodies are in motion towards oppowebsite directions, the family member speed is calculated by summing the rate of both. Consider two objects moving through speeds of and in oppowebsite directions. Here the family member rate is x+y.

The loved one rate differs via the family member velocity by a major aspect. The distinction between the two is that relative speed is a scalar amount , that suggests has only magnitude whereasloved one velocity is a vector quantity which has both magnitude

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## What you'll learn:

## Illustration of Relative Speed

How have the right to we calculate the relative speed?

Consider an example. Suppose 2 objects A and also B are relocating at sepaprice speed in the exact same direction.

Let the first object’s speed be 9 km/hr and the second object’s rate is 5 km/hr.

Their family member speed will be,

=(9−−5)km/hr = left( 9 m -- m 5 ight) m km/hr;=(9−−5)km/hr=4km/hr <9>5> = m 4 m km/hr;left< 9 m > m 5 ight>=4km/hr<9>5>

Then, The time taken for both the objects to satisfy

=distancecoveredrelativespeed =frac m distance extended m family member speed=relativespeeddistancecovered=dkm(9−−5)km/hr=fracd m kmleft( 9 m -- m 5 ight) m km/hr=(9−−5)km/hrdkm

## Instances for Relative Speed

Example no. 1:

One robber is noticed by a policeguy who is standing at a Distance of 300m. They began running after seeing each various other. What is the distance covered by the robber running at 7 kmph before the policemales running at 10 kmph catches him?

Solution:

Here two world moving at sepaprice Speeds.

So, let us assume robber to be stationary. Take the speed of Policeguys via respect to robber.

So, by utilizing the principle of family member speed, we can note that robber is standing and also policemen at 300 m is running at 10−7=3kmph.10 m - m 7 m = m3 m kmph.10−7=3kmph.

Thus, to capture the robber he needs to take a trip 0.3 km.

Timetaken=0.3/3=0.1hours mTime m taken m = m 0.3/3 m = m 0.1 m hoursTimetaken=0.3/3=0.1hours

So, the Distance spanned by robber before policemen catches him is 0.7 km.

Example 2:

A student rowed a boat at a rate of 16 kmph upstream and 22 kmph downstream. Find theSpeed of the stream and discover the Speed via which the student rows the boat in still water.

Givenspeedofupstream=16kmph mGiven rate of upstream = 16 kmphGivenspeedofupstream=16kmphSpeedofDownstream=22kmph mSpeed of Downstream = 22 kmphSpeedofDownstream=22kmphStudent′sspeedinstillwater=x mStudent's speed in still water = xStudent′sspeedinstillwater=x=(a+b)2 = fracleft( a + b ight)2=2(a+b)=(16+22)2 = fracleft( 16 + 22 ight)2=2(16+22)=382 = frac382=238=19kmph = 19 m kmph=19kmphSpeedofthestream=y mSpeed of the stream = ySpeedofthestream=y=(a−b)2 = fracleft( a - b ight)2=2(a−b)=(16−22)2 = fracleft( 16 - 22 ight)2=2(16−22)=−62 = frac - 62=2−6=3kmph = 3 m kmph=3kmph

Example 3:

Speed of a River is 8 kmph. A stationary object is preserved in the river. Find the Time taken by the floating object to reach a rock which is 16 kilometres downstream from the location wright here it is now?

solution:

speedoftheobject=speedofriver=8kmph mspeed of the object = rate of river = 8 kmphspeedoftheobject=speedofriver=8kmphspeed=distancetime mspeed = frac mdistance mtimespeed=timedistanceso,timetakentoreach16km=168=2hours mso, time taken to reach 16 kilometres = frac168 = 2 m hoursso,timetakentoreach16km=816=2hours

Example 4:

Find the Time taken for a 140m lengthy train moving at a speed of 10m/s to overtake anothertrain which is 100m lengthy and moving in the very same direction at a Speed of 6 m/s?

Solution:

Let us Suppose that 100m lengthy train is stationary and taking the rate of the 140m long train via respect to the previous

So, the loved one Speed of 140m long train = 10 - 6 = 4 m/s

To overtake the 100m lengthy train it need to cover a complete Distance = sum of its very own size and also other train’s size = 140 m + 100 m = 240 m

So, to travel that Distance via a Speed of 6 m/s, it will certainly take 2406=40secfrac2406 = 40 m sec6240=40sec

Example 5:

Two cars are travelling from the same area at the speed of 15 km/hr and 10 km/hr respectively. Find the distance between the cars after 8 minutes offered that both cars are moving in the same direction.

See more: In Each Case Determine The Value Of The Constant C, Answered: In Each Case, Determine The Value Of…

Solution:

(15−−10)km/hrleft( 15 m -- m 10 ight) m km/hr m (15−−10)km/hr=5km/hr = m 5 m km/hr=5km/hr

speed×time=(5×8/60)kmtablespeed m imes m time = m left( 5 m imes m 8/60 ight) m km m table speed×time=(5×8/60)kmtable=2/3km=2/3×1000m=666.66m = m 2/3 m kilometres = m 2/3 m imes m 1000 m m m = m 666.66 m m=2/3km=2/3×1000m=666.66m

Therefore, the distance in between the cars after 8minutes is 666.66 m offered that 2 cars are travelling in the exact same direction.

Example 6:

A student A starts at 10AM and starts running at a rate of 20kmph. At 11AM student B starts chasing student A at a speed of 30kmph. Find at what time Student B overtake Student A?

Solution:

If we compare the times of student A and B, Student B has actually a lag of one hour

That is 11-10 = 1 hour

Student B chases Student A which suggests it is relocating in the exact same direction as Student A at a speed of 30 kmph

So, Student B was faster, and Student A is slower

When two objects are moving in the exact same direction, the relative speed will be the rate of much faster object minus rate of sreduced object

Speed of student B - rate of student A = 30-20 = 10 kmph

With each passing hour Student B gains 10kmSo, after three hrs Student A cover 60 km (20 x 3 = 60)