A girl travels along a straight line, from point A to B at a constant speed, V1 meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V2 meters/sec for another T seconds. BC makes an angle 105° with AB. If CA makes an angle 30° with BC, how much time will she take to travel back from point C to A at a constant speed of V2 meters/sec, if she travels along a straight line from C to A?
Explanation:
Refer to the diagram below.
Distance AB = V1T and distance BC = V2T
Also, ∠CAB = 180 – 30 – 105 = 45°
Let BD be drawn perpendicular to AC.
In ΔBDA, BD = AB sin 45° = V1T/√2 Also, AD = AB cos 45° = V1T/√2
In ΔBDC, BD = BC sin 30° = V2T/2 Also, DC = BC cos 30° = √3V2T/2
Now, V1T/√2 = V2T/2 ⇒ V1 = V2/√2
AC = AD + DC = V1T/√2 + √3V2T/2 = V2T/2 + √3V2T/2 = V2T(1 + √3)/2
Time taken to travel AC at speed V2 = [V2T(1 + √3)/2] ÷ V2 = T(1 + √3)/2 = 0.5(1 + √3)T
Hence, option (c).
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