Concept:This problem involves vector displacement on a plane, where the distance between two parallel streets is found by adding the eastward components of two travel legs.
Explanation:The person starts on the first street, facing North. He wants to reach the second street on his right (East).
He first makes a right turn of
150∘ from North, so his direction becomes
150∘ clockwise from North, which is
60∘ South of East.
He travels for
15 minutes at
20 km/hr. Distance covered =
20×6015=5 km.
The eastward component of this leg is
5×sin150∘=5×21=2.5 km.
Next, he takes a left turn of
60∘ from his current direction (
150∘). The new direction is
150∘−60∘=90∘, i.e., due East.
He travels for
20 minutes at
30 km/hr. Distance covered =
30×6020=10 km, entirely eastward.
Total eastward displacement from the starting point =
2.5+10=12.5 km.
Since the streets are parallel and North-South, the perpendicular distance between them equals the total eastward displacement.
Therefore, the distance between the two streets is
12.5 km.
Answer:12.5 km (Option C)
