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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?
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
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
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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!!!!!!!!!!!!!!!!!!!!!!!!!!
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.
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