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neha shugani Grade: 12
        

A particle moving in a straight line covers half the distance with speed of 3m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is?

6 years ago

Answers : (8)

Shivam Bhagat
22 Points
										Let total distance=d

Total time=T
Then in d/2.......
t1=d/6

t2=d/(4*4.5) + d/(4*7.5)
=4d/45

T=t1+t2=d/6+4d/45 =23d/90

Average speed =t. Dis/t. Time= d/T =3.91
m/s........ANS
All the best!!!!!
6 years ago
vikas askiitian expert
510 Points
										

let total distance of path is S ...


 


for first half , time taken if t


speed = distance/time


3 = S/2t


t = S/6      .............1


 


for second half , time taken is T ...


for first  T/2 it moves with 4.5


speed = distance/time


4.5 = 2x/T


 x = 4.5T/2   ...........2


for second T/2 it moves with 7.5


7.5 = 2y/T


y = 7.5T/2       ..........3


we have , x+y = S/2so


(7.5+4.5)T = S/2


T = S/24        ...............4


total time taken during its motion is T+t


 Ttotal = T+t = S/24+S/6 = 5S/24


total distance = S


average speed = total distance/total time = S/5S/24 = 24/5 = 4.8m/s


 


approve if u like my ans

6 years ago
iit jee
44 Points
										

hi


let the distanc covered with speed 3m/s be x.mts


time taken is x/3sec.


ii)second half(x mts)


time taken to travel second half will be 18x/11.25 sec


avg. speed=total distanc /totaltime


avg.speed=1.9m/s


plzz. prove!!!!!!!!!!!!!!!!!!!!!!!!!!

6 years ago
Himanshu Dogra
34 Points
										

Let total distance be S. Total time be T and V1= 3m/s be the velocity for which it covers the first half of the distance and V2 = 4.5 m/s and V3 = 7.5 m/s be the velocities in the second half of the distance. Let <v> be the average velocity that has to be calculated.


Let the other half be completed in times t1 and t2 (t1= t2=t, from the question).


For Ist half of the distance:


S/2= V1(T-2t) 


=> T = S/2V1 + 2t              ....(1)


For 2nd half of the distance:


S/2 = (V2 + V3)2t


=> S = 4(V2 + V3)t             ....(2)


From (1) and (2).


T = {2(V2 + V3)/V1    + 2}t      ....(3)


we know that,


<v> = total distance/total time


so,


<v> = S/T


Putting the values of  T and S from (3) and (2),


<v> = 4(V2 + V2)t /{2(V2 + V3)/V1   +2 }t


=> <v> = 2V1 (V2 + V3)/(2V1 + V2 + V3)


putting the values of V1, V2 and V3.


<v> = 4 m/s.

6 years ago
vijval
14 Points
										
all the answers are waste of time 
there is a formulae 2v1*v2/v1+v2 if the particle travels in 2 equal time intervals so lets take second distance 4.5+7.5/2 =12m/s so 2*v1*v2/v1+v2 = 2*3*6/6+3 =36/9=4.0 m/s 
11 months ago
vijval
14 Points
										
all the answers are waste of time 
there is a formulae 2v1*v2/v1+v2 if the particle travels in 2 equal time intervals so lets take second distance 4.5+7.5/2 =12m/s so 2*v1*v2/v1+v2 = 2*3*6/6+3 =36/9=4.0 m/s 
answered by a 8th grade
11 months ago
Ashutosh
21 Points
										Avg.speed =total distance /total time taken Avg. Speed in 2nd half=(sT +ST)/2T                               = (4.5T+7.5T)/2T=6m/sTotal avg.Speed =  D/(d/2)/3+(d/2)/6                        =12D/3D=4m/s
										
4 months ago
Abhijith baiju
19 Points
										V1=3m/sV2=4.5 m/sV3=7.5 m/sTherefore formula for dist covered in unequal speed in equal intervals of time=2V1(V1+V3)/V1+V2+V3=2x3(4.5+7.5)/3+4.5+7.5=4.875
										
23 days ago
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