Question

Joseph jogs from one end A to the other end B of a straight 300m road in 2 minutes 30 seconds and turns around and jogs 100m back to point C another 1 minute. What are Joseph’s average speeds and velocities in jogging
(a) from A to B
(b) from A to C

Answer 1

Average speed of a body is defined as the total distance covered by the body divided by the total time taken by the body to cover this distance. Average velocity of a body is defined as the net displacement of the body divided by the total time taken by the body.

Complete answer:
Let us first understand what is meant by average speed and average velocity.
Average speed of a body is defined as the total distance covered by the body divided by the total time taken by the body to cover this distance.
Average velocity of a body is defined as the net displacement of the body divided by the total time taken by the body.
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to $150\text{seconds}$. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of $150\text{seconds}$.Hence, his average speed is $\dfrac{300m}{150s}=2m{{s}^{-1}}$.Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is $\dfrac{300m}{150s}=2m{{s}^{-1}}$.

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of $300m+100m=400m$. And he covers this total distance in a time interval of $2.5\min +1\min =3.5\min =210\text{s}$.
Therefore, his average speed for this journey is $\dfrac{400m}{210s}=1.9m{{s}^{-1}}$.
For the same journey is displacement is equal to the distance between the points A and C,i.e. $300m-100m=200m$.

Hence, his average velocity for this case is $\dfrac{200m}{210s}=0.95m{{s}^{-1}}$.

Note: The distance covered by a body is the total length of the route by which the body travels. It is just a magnitude.Displacement of a body is a vector joining the initial position and the final position of the body. Its magnitude is equal to the length of the line segment joining the initial and final positions.