Question

On a long horizontally moving belt shown in figure, a child runs to and fro with a speed 9 km h⁻¹ (with respect to the belt) between his father and mother located 50 metres apart on the moving belt. The belt moves with a speed of 4 km h⁻¹. For an observer on a stationary platform outside, what is the
a. speed of the child running in the direction of motion of the belt?
b. speed of the child running opposite to the direction of motion of the belt?
c. time taken by the child in (a) and (b)
Which of the answers alter if motion is viewed by one of the parents?

Answer 1

Hey buddy,

# Answers-
a) Child runs in direction of belt,
vcb = 3.61 km/h
t = 13.9 s
b) Child runs in opposite direction of belt,
vcb = 1.39 m/s
t = 36 s

# Given-
Speed of child vc = 9km/h = 2.5 m/s
Speed of belt vb = 4km/h = 1.11 m/s

# Solution-
a) When the child runs in direction of belt
vcb = vc-(-vb)
vcb = 2.5-(-1.11)
vcb = 3.61 m/s
Also
t = s/vcb
t = 50/3.61
t = 13.9 s
Hence, acording to observer, v=3.61m/s and t=13.9s.

b) When the child runs in opposite direction of belt
vcb = vc-vb
vcb = 2.5-1.11
vcb = 1.39 m/s
Also
t = s/vcb
t = 50/1.39
t = 36 s
Hence, acording to observer, v=1.39m/s and t=36s.

Hope that solved your problem...

Answer 2

Let us consider the left to right to be positive direction of x axis:

a) Velocity of belt = 4 km/he

Speed of child  w.r.t belt = vc=9 m/h = 9x 5/18 =5/2 m/s

Speed of child w.r.t stationary observer  =Vc'= 9+4=13 km/h

b) Speed of belt=vb= 4km/h

vc= - 9km/h

speed of child w.r.t stationary observer =vc'= vc+vb

=- 9 +4 = - 5km/h

Negative direction implies that child runs in the opposite direction of belt .

c) Distance between the parents=540 km

we can see that both parents and child are in same  belt .

so speed of child  

observed by stationary observer in either direction will be 9km/h

t= 50/ (5/2)

=20 s