In a triangle ABC, AB=BC=CA=30 m. A man covers the edges of the triangle by running with different velocities such that, Velocity from B to C = 10 m/s, velocity from C to A = 20 m/s and velocity from A to B= 30 m/s. What is the average speed of this man? {Answer :16.36 m/s.} show steps how it comes out.
Answers
Answer:
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Explanation:
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motion of the particles is roughly sketched in Fig. By symmetry they will meet at the centroid O of the triangle. As the speed of all the particles is equal they will cover equal distance in any given interval of time.
If we join the instantaneous position of the particle at any time, the particles will form an equilateral triangle of same centroid as initial triangle.
Let us consider the motion of any one particle say A. At any instant, its velocity makes an angle 30
o
with line AO.
The component of the velocity along AO is vcos30
o
. This component will be equal to the rate of change of distance between A and O.
At t=0, distance between A and O,
AO=
3
2
AD=
3
2
l
2
−(
2
l
)
2
=
3
l
At time t=T, the separation between A and O is zero. Hence, time taken for AO to become zero.
T=
vcos30
o
(l/
3
)
=
v(
3
/2)
(l/
3
)
=
3v
2d
After time t, let the distance of separation between the insect be r. The velocity of approach (component of relative velocity v
rel
between the insects along the line of their separation) is = v+vcos60
o
=3v/3
Since (v
rel
)
II
dt=−dr substituting (v
rel
)
II
=
2
3v
, we obtain
2
3v
dt=−dr
As at time of meeting, integrating both sides, we have
2
3v
∫
0
τ
dt=−∫
l
0
dr
This gives t=
3v
2l
solution