Question 203: Motion in 2D & Vectors

Question

Two boys are standing at the ends A and  B of a ground where AB = a. The boy  at B starts running in a direction  perpendicular to AB with velocity v
1
. The boy at A starts running  simultaneously with velocity v and catches the other in a time t, where t is

Accepted Answer

Velocity of A relative to B is given by
$$\overrightarrow {{v_{A/B}}}  = \overrightarrow {{v_A}}  - \overrightarrow {{v_B}}  = \overrightarrow v  - \overrightarrow {{v_1}} $$    ...(i)
By taking x-components of equation (i), we get
$$0 = v\sin 	heta  - {v_1} \Rightarrow \sin 	heta  = {{{v_1}} \over v}$$    ...(ii)
By taking Y-components of equation (i), we get
$${v_y} = v\cos 	heta $$    ...(iii)
Time taken by boy at A to catch the boy at B is
given by
t = $${{{\mathop{m Relative}
olimits} \,displacement\,along\,Y - axis} \over {{\mathop{m Relative}
olimits} \,velocity\,along\,Y - axis}}$$
= $${a \over {v\cos 	heta }} = {a \over {v.\sqrt {1 - {{\sin }^2}	heta } }} = {a \over {v.\sqrt {1 - {{\left( {{{{v_1}} \over v}} ight)}^2}} }}$$
[From equation (i)]
= $${a \over {v.\sqrt {{{{v^2} - v_1^2} \over {{v^2}}}} }} = {a \over {\sqrt {{v^2} - v_1^2} }}$$
= $$\sqrt {{{{a^2}} \over {{v^2} - v_1^2}}} $$

Answer

$${a \over {\sqrt {{v^2} + {v_1}^2} }}$$

Answer

$${a \over {v + {v_1}}}$$

Answer

$${a \over {v - {v_1}}}$$

Answer

$$\sqrt {{a \over {{v^2} - {v_1}^2}}} $$

Motion in 2D & Vectors

Two boys are standing at the ends A and B of a ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v 1 . The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

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