Question

A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same direction. What is the largest distance the student could possibly be from the starting point?

Answer 1

Answer:

7.1 m

Explanation:

Given:

Distance traveled by the student in the first attempt = $(2.9 \pm 0.1)\ m$

Distance traveled by the student in the second attempt = $(3.9 \pm 0.2)\ m$

So, the maximum distance that the student could travel in this attempt = $(2.9+0.1)\ m = 3.0\ m$

So, the maximum distance that the student could travel in this attempt = $(3.9+0.2)\ m = 4.1\ m$

Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.

So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.

And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.

Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.