An airplane is observed by two observers traveling at 60 km h^-1 in two vehicles moving in opposite directions on a straight road. To an observer in one vehicle, the plane appears to cross the road track at right angles while to the other appears to be 45^@. At what angle does the plane actually cross the road track and what is its speed relative to ground ?

An airplane is observed by two observers traveling at 60 km h^-1 in tw

1.	An airplane is observed by two observers traveling at 60 km h^-1 in two vehicles moving in opposite directions on a straight road. To an observer in one vehicle, the plane appears to cross the road track at right angles while to the other appears to be 45^@. At what angle does the plane actually cross the road track and what is its speed relative to ground ?
Answer» Solution :Let `v_p` be the velocity of plane relative to the ground, at angle `theta` to velocity `vec v_1` of observer in car `1`. In case (i), `vec v_(P_1) = vec v_p - vec v_1` `vec v_P = vec v_(P_1) + vec v_1` Vector diagram is shown in (Fig. S5.46). Note that according to observer in car `1`, the plane crosses the road at right ANGLES. Similarly, in case `2 vec v_P = vec v_(p_2) + vec v_2` We can combine (Figs S5.45) (a) and (S5.45) (B), `tan 45^@ = (AC)/(AB)` `v_(P_1) = (v_1 + v_2) tan 45^@ = 120 xx 1 = 120 km h^-1` `v_p = [60^2 +120]^(1//2) = 134.16 km h^-1` `tan theta = (v_(P_1))/(v_1) = (120)/(60) = 2` Hence, `theta = tan^-12`. , .