On a long horizontally moving belt, a child runs to
and fro with a speed 9 kmh (with respect to the belt)
between his father and mother located 50 m apart
on the moving belt. The belt moves with a speed of:
4 km h ! For an observer on a stationary platform,
the speed of the child running in the direction of motion
of the belt is
Answers
Explanation:
(a) Speed of the belt, vB = 4 km/h
Speed of the boy, vb = 9 km/h
Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + vB = 9 + 4 = 13 km/h
(b) Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + (– vB) = 9 – 4 = 5 km/h
(c) Distance between the child’s parents = 50 m
As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.
Hence, the time taken by the child to move towards one of his parents is 50 / 2.5 = 20 s
(d) If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.
For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.
Answer:
Explanation:
(a) Given ,
Speed of belt = 4 km/h
speed of child with respect to belt (Vcb) = 9 km/h
Vc - Vb = 9 km/h
Vc = 9km/h + Vb
= 9km/h + 4km/h
= 13km/h
Hence, child runs in the direction of motion of the belt = Vc = 13 km/h
(b) when child runs in the opposite direction of the belt
Vb = 4 km/h
Vcb = -9km/h
Vc - Vb = -9km/h
Vc = -5 km/h
Negative sign indicates that child moves opposite direction of the motion of the belt .
(c) the speed of child observed by stationary observer in both condition is same due to child and parants are present in same belt .{ only change it direction e.g in the direction 9 km/h and opposite the direction -9km/h }
Hence, time taken by child in both cases = distance/speed
= 0.05/9 hour
= 20 sec
If motion of child observed by one parants then answer of case (a) and (b) both are change but answer of( c ) always remains same becoz child and parents are in same belt .