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
550.1k+ views
Hint: 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.
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.
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