Question : A particle moves half the time of its journey with velocity 20 km/h. The rest half time it moves with two velocities such that half the distance it covers with 30 km/h and the other half with 60 km/h. Find the net average velocity assume straight line motion (in km/h).
Doubt by Suhani
Solution :
Let the total time taken by the particle to cover the journey = t hours
Distance covered by the particle in first half time (t/2) with velocity 20 km/h
We know,
Speed = Distance / Time
Distance = Speed × time
x1= 20 × (t/2)
x1 = 10t — (1)
Total Distance (x2) covered by the particle in the second half time (t/2) with velocity 30 km/h in the first half distance and 60 km/h in the second half distance.
We know,
Speed = Distance / Time
Speed = Distance / Time
Time = Distance / Speed
t/2 = [x2/2]/30 + [x2/2]/60
t/2 = x2/60 + x2/120
t/2 = 2x2/120 + x2/120
t/2 = 3x2/120
t/2=x2/40
40t/2=x2
20t=x2
t/2 = [x2/2]/30 + [x2/2]/60
t/2 = x2/60 + x2/120
t/2 = 2x2/120 + x2/120
t/2 = 3x2/120
t/2=x2/40
40t/2=x2
20t=x2
x2=20t — (2)
Now,
Average Velocity
Average Velocity
= Total Distance Covered / Total Time Taken
= [x1+x2]/t
= [10t+20t]/t
= [10+20]t/t
= 30 km/h
= [x1+x2]/t
= [10t+20t]/t
= [10+20]t/t
= 30 km/h
Hence, the net average velocity will be 30 km/h.
Similar Question :
1.) A person move in such a way that he travel first half time of his journey with speed 36 km/h and in remaining half time he cover first half distance with speed of 20 km/h and remaining half distance with speed 30 km/hr. Find the average speed?
a) 30 km/h
b) 24 km/h
c) 27 km/h
d) 48 km/h
2.) A particle travels half of the time with constant speed of 2m/s, in remaining half of the time it travels, (1/4)th distance with constant speed of 4 m/s and (3/4)th distance with 6 m/s. Find average speed during the complete journey.
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1.) a) 30 km/h2.) 11/3 m/s