Question

three elephants A,B,C are moving along a straight line with constant speed in same direction as shown in figure. Speed of A is 5 m/s and speed of C is 10 m/s . Initial 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. find speed of B.

Answer 1

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Answer 2

"

The speed of B elephant is 15m/s.

Given:

Speed of A is 5 m/s

Speed of B is 10 m/s

Solution:

Let us consider the points A and C

The initial displacement between AC is equal to 2D

The final displacement between AC is equal to 3D

Therefore,

Net displacement is 3D-2D = 1D

Relative velocity $= V_{C}-V_{A}=5 \mathrm{m} / \mathrm{s}$

We know that, $Velocity\quad = \quad \frac { Displacement }{ Time }$

Therefore, time taken for B and C to meet

$V_{ S }-V_{ A }\quad =\quad \frac { net\quad displacement }{ time\quad (t) } 5\quad =\quad \frac { D }{ t }$

Thus, $t\quad =\quad \frac { D }{ 5 }$... consider as eqn(1)

Now, considering the points A and B

Initial displacement between A and B is equal to D

Final displacement between A and B is equal to 3D

Total displacement between A and B = 3D - D = 2D

By Using eqn (1) we have $time (t) = \frac{D}{5}$

Thus,

$V_{ AB }\quad =\quad \frac { { Total\quad displacement\quad between\quad }A\quad { and }\quad B }{ { time }(t) }$

$V_{ AB }\quad =\quad \frac { 2D }{ \left( \frac { D }{ 5 } \right) } \quad =\quad \frac { 2D\quad \times \quad 5 }{ D } \quad =\quad 10\quad \frac { m }{ s }$

Therefore, $V_B = 10 + V_A$

= 10 + 5

$V_B = 15 m/s$"