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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? 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?
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=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!!!!!
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
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
approve if u like my ans
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!!!!!!!!!!!!!!!!!!!!!!!!!!
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.
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.
all the answers are waste of timeย 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+v2ย = 2*3*6/6+3 =36/9=4.0 m/sย
all the answers are waste of timeย 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+v2ย = 2*3*6/6+3 =36/9=4.0 m/sย answered by a 8thย grade
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
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
Average Speed (half distance formulae):=>2ร[V1รV2] / V1+V2.=>[3ร(4.5+7.5)] / 6+3=>[3ร12]/9=>36/9=>4.0 m/s.๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐
Hello StudentLet the total distance covered be 2x.Let t1 be the time taken the first half of the distance i.e. x.Let the other half of the distance (x) be covered in two equal intervals of time t2.x = 4.5t2 + 7.5t2x = 12t2t2 = x/12Total Time (T) = t1+ 2t2= x/3 + x/6 = x/2Average Speed = total distance travelled / total time takenAverage speed = (2x)/x/2 = 4ms-1.I hope this answer will help you.
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