Question : A body is travelling along a straight line traversed one third of the total distance with a velocity of 4 m/s the remaining part of the distance was covered with the velocity of 2 m/s for half of the time and with velocity 6 m/s for other half of time. What is the mean velocity averaged over the whole time of motion?
Doubt by Tanay
Solution :
Let the total distance covered is s.
Case I
Distance Covered (s1) = s/3
Velocity (v1) = 4 m/s
Time (t1) = Distance/speed
t1=(s1)/v1
t1=(s/3)/4
t1=s/12
t1=(s/3)/4
t1=s/12
Case II
Distance covered (s2)=[s-(s/3)]
s2=(2s/3)
s2=2s/3
s2=(2s/3)
s2=2s/3
t2=t3
We know
We know
Speed = Distance / time
Distance = Speed × time
2s/3 = v2×t2+v3×t3
2s/3=2×t2+6×t3
2s/3=2×t2+6×t2 [∵t2=t3]
2s/3=t2(2+6)
2s/3=8t2
2s/(3×8)=t2
2s/24=t2
s/12=t2
Hence,
Distance = Speed × time
2s/3 = v2×t2+v3×t3
2s/3=2×t2+6×t3
2s/3=2×t2+6×t2 [∵t2=t3]
2s/3=t2(2+6)
2s/3=8t2
2s/(3×8)=t2
2s/24=t2
s/12=t2
Hence,
t2=t3=s/12
Now,
Average velocity (mean velocity)
= (Total distance covered) / (Total time taken)
= (s1+s2+s3) / (t1+t2+t3)
= [(s/3)+(s/3)+(s/3)]/[t1+t2+t3]
= s/[(s/12)+(s/12)+(s/12)]
=S/[3s/12]
=1/[1/4]
=4/1
=4 m/s
Hence, 4m/s is the mean velocity averaged over the whole time of motion.
= s/[(s/12)+(s/12)+(s/12)]
=S/[3s/12]
=1/[1/4]
=4/1
=4 m/s
Hence, 4m/s is the mean velocity averaged over the whole time of motion.