Motion in a Straight Line
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Question : 9
Total: 27
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h– 1
Solution:
Let the speed of each bus = v b k m h − 1
= v c = 20 k m h − 1
Case I : Relative speed of the buses plying in the direction of motion ofcyclisti . e . A B = v b − v c = ( v b − 20 ) k m h − 1
since the bus goes past the cyclist every 18 minutes( =
h )
∴ = ( v b − 20 ) ×
k m . . . . . . .
Since the bus leaves after every T minutes, so the distance covered by thebus in T min ( =
h )
= v b ×
. . . . . . .
∴ From (i) and (ii), we get
or( v b − 20 ) ×
= v b ×
orv b − 20 = v b ×
. . . . .
Case II : Relative speed of the bus coming from town B to A w.r.t. cyclist( v b + 20 ) k m h – 1
Since the bus goes past the cyclist after every 6 minutes
∴ Distance covered by the bus w.r.t. the cyclist
= ( v b + 20 ) ×
k m . . . .
Also distance covered by bus in T minutes is
= v b ×
k m . . . .
∴ Equating (iv) and (v), we get
( v b + 20 ) ×
= v b ×
orv b + 20 = v b ×
. . .
Dividing (vi) by (iii), we get
v b + 20 v b − 20 = 3
or v b + 20 = 3 v b − 60
or2 v b = 80
orv b = 40 k m h − 1
Putting the value ofv b
40 − 20 = 40 ×
or20 = 40 ×
or T = 20 ×
= 9
∴ v b = 40 k m h − 1 ,
T = 9
Case I : Relative speed of the buses plying in the direction of motion ofcyclist
since the bus goes past the cyclist every 18 minutes
Since the bus leaves after every T minutes, so the distance covered by thebus in T min
∴ From (i) and (ii), we get
or
or
Case II : Relative speed of the bus coming from town B to A w.r.t. cyclist
Since the bus goes past the cyclist after every 6 minutes
∴ Distance covered by the bus w.r.t. the cyclist
Also distance covered by bus in T minutes is
∴ Equating (iv) and (v), we get
or
Dividing (vi) by (iii), we get
or
or
or
Putting the value of
or
or
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