Question

The IIT lecture theatre is 50 ft wide and has a door at the corner. A teacher enters at 7.30 AM through the door and makes 20 rounds along the 50 ft wall back and forth during that period and finally leaves the class-room at 9 AM through the same door. If he walks with constant speed of 10 ft/min, find the distance and the magnitude of displacement travelled by the teacher during the period:
A) 2000 ft and 0 ft
B) 100 ft and 0 ft
C) 2000 ft and 50 ft
D) None of these

Answer 1

Hint
This is a pretty straightforward question; all we need to keep in mind is that distance and displacement are 2 different things and are independent of the time taken. Distance is the total length covered in the motion and the displacement is the difference between two points in the path of the motion. Since the motion in the room started and ended at the same point, the displacement is zero. But the distance depends on the path taken.

Complete step by step answer
The distance travelled by a body in time $t$ is the total length covered by it. In this question, it is given that the teacher walks along the $50 ft$ wall back and forth, so the distance travelled by him in 1 round will be:
$s\,\, = \,50\, \times \,2\, = \,100ft$
As he makes 20 rounds of along the same line, the total distance travelled by him in from 7.30 to 9 AM is:
$s\, = \,100\, \times 20\, = \,2000\,ft$
Therefore, the total distance travelled by him is equal to $2000 ft$ and hence option (B) is strike off. The displacement of the body in time t is given as the difference between the final and initial positions of the body. In our case, the teacher returns to his original state after the class and thus his difference in initial and final position is $0$. Therefore, his displacement is also $0$.
So, the correct answer is option A.

Note
The only difference between distance and displacement is that one is scalar and the other is a vector quantity i.e. the value of displacement can be 0 and even by negative depending on its direction of motion.