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Question: Three elephants A, B and C are moving along a straight line with constant speed in the same directio...

Three elephants A, B and C are moving along a straight line with constant speed in the same direction as shown in figure. Speed of A is 5m/s5m/s and speed of C is 10m/s10m/s . Initially separation between A and B is d and between B and C is also d .when B catches C separation between A and C becomes 3d. Then the speed of B will be:


A) 15m/s15m/s
B) 7.5m/s7.5m/s
C) 20m/s20m/s
D) 5m/s5m/s

Explanation

Solution

We can solve this question by using the relative velocity concept. First we calculate what time distance becomes 3d3d between A and C.
In the same time B catches C it is given then we calculate in what time B catches C and equate the time.

Complete step by step solution:
Step 1
First we calculate in what time separation between A and C become 3d3d
Elephant A and B having initial separation 2d2d we assume after time tt separation become 3d3d means separation increased by dd this distance dd will be travel by elephant C with relative velocity with respect elephant A
Relative velocity of elephant C with respect A is VCA{V_{CA}} can be given as
VCA=VCVA\Rightarrow {V_{CA}} = {V_C} - {V_A}
In question given VC=10m/s{V_C} = 10m/s and VA=5m/s{V_A} = 5m/s put these in above equation
VCA=105 VCA=5m/s  \Rightarrow {V_{CA}} = 10 - 5 \\\ \Rightarrow {V_{CA}} = 5m/s \\\
Hence dd distance travel in time tt with velocity VCA{V_{CA}}
We know
velocity=distance x time
VCA=dt 5=dt  \Rightarrow {V_{CA}} = \dfrac{d}{t} \\\ \Rightarrow 5 = \dfrac{d}{t} \\\
So time
t=d5\therefore t = \dfrac{d}{5} ........ (1)

Step 2
Elephant B catches C in same time in which time separation increase by dd between A and C
Elephant B catches C in time tt so
Relative velocity of elephant B with respect to C is
VBC=VBVC\Rightarrow {V_{BC}} = {V_B} - {V_C}
VBC=VB10\Rightarrow {V_{BC}} = {V_B} - 10
With this relative velocity B travel d distance in time tt
Apply velocity=distance x time
VBC=dt\Rightarrow {V_{BC}} = \dfrac{d}{t}
VB10=dt\Rightarrow {V_B} - 10 = \dfrac{d}{t}
So time
t=dVB10\therefore t = \dfrac{d}{{{V_B} - 10}} ........ (2)
In question it is given that at the same time B catches C it is given then we calculate in what time B catches C and equate the time.
From equation (1) and (2)
(1)= (2)
d5=dVB10\Rightarrow \dfrac{d}{5} = \dfrac{d}{{{V_B} - 10}}
Solving this
VB10=5 VB=5+10  \Rightarrow {V_B} - 10 = 5 \\\ \Rightarrow {V_B} = 5 + 10 \\\
VB=15m/s\therefore {V_B} = 15m/s
Hence velocity of elephant B is 15m/s15m/s

Hence option A is correct

Note: Here we use relative velocity concept which can be understand as
VAB=VAGVBG{V_{AB}} = {V_{AG}} - {V_{BG}}
Where VAB{V_{AB}} \Rightarrow Relative velocity of A with respect to B
VAG{V_{AG}} \Rightarrow Velocity of A with respect to ground
VBG{V_{BG}} \Rightarrow Velocity of B with respect to ground