Answer
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Hint: Average speed is the total distance of a body divided by the total time taken by it. Average velocity of the body is the net displacement of the body divided by the taken time. Use these formulas to find the average speed and velocities of both the vehicles.
Formula Used:
$u=\dfrac{d}{t}$
$v=\dfrac{s}{t}$
Complete step-by-step answer:
There are two bodies given in the question, the aeroplane and the deluxe bus. The aeroplane travels from Kota to Jaipur and the deluxe bus travels from Jaipur to Kota.
When we travel by air there is no traffic as that of while traveling by road. Therefore, the aeroplane will cover the shortest distance.
It is given that the aeroplane covers a distance of 260 km in 15min and a deluxe bus covers a distance of 320 km in 8 hours.
(i) Let us calculate the speed of the bus.
Average speed of a body is defined as the total distance by that body divided by the total time taken to cover this distance.
Let the average speed of the bus be ${{u}_{B}}$. Let the distance covered by the bus be ${{d}_{B}}$ and time taken be ${{t}_{B}}$.
Hence, the average speed of the bus will be ${{u}_{B}}=\dfrac{{{d}_{B}}}{{{t}_{B}}}$ ….. (i).
It is given that ${{d}_{B}}=320km$ and ${{t}_{B}}=8h$
Substitute the values of ${{d}_{B}}$ and ${{t}_{B}}$ in equations (i).
$\Rightarrow {{u}_{B}}=\dfrac{320}{8}=40km{{h}^{-1}}$
Hence, the average speed of the bus is 40km/h.
(ii) Let us calculate the average velocity of the bus.
Average velocity of a body is defined as the net displacement divided total time taken.
Let the average velocity of the bus be ${{v}_{B}}$. Let the distance covered by the bus be ${{s}_{B}}$ and time taken be ${{t}_{B}}$.
Then the average velocity of the bus is ${{v}_{B}}=\dfrac{{{s}_{B}}}{{{t}_{B}}}$ ….. (ii).
Let us consider the direction from Jaipur to Kota. The magnitude of net displacement of the bus is the length of line segment joining the two cities and it is the shortest distance between the two cities. We know that the aeroplane covers the shortest distance between the two cities. Hence, the displacement of the bus is 320km. ${{s}_{B}}=260km$.
It is given that ${{t}_{B}}=8h$.
Substitute the values in equation (ii).
$\Rightarrow {{v}_{B}}=\dfrac{260}{8}=32.5km{{h}^{-1}}$.
This means that the average velocity of the bus is 32.5km/h.
(iii) Let us calculate the average speed of the aeroplane. We already know what the average speed of a body is.
Let the average speed of the aeroplane be ${{u}_{A}}$. Let the distance covered by the aeroplane be ${{d}_{A}}$ and time taken be ${{t}_{A}}$.
$\Rightarrow {{u}_{A}}=\dfrac{{{d}_{A}}}{{{t}_{A}}}$ …. (iii).
It is given ${{d}_{A}}$=260km and ${{t}_{A}}=15\min =15\times \dfrac{1}{60}h=0.25h$
Substitute the values in equation (iii).
${{u}_{A}}=\dfrac{260}{0.25}=1040km{{h}^{-1}}$.
Hence, the average speed of the aeroplane is 1040km/h.
(iv) Let us calculate the average velocity of the aeroplane.
Let the average velocity of the bus be ${{v}_{A}}$. Let the distance covered by the bus be ${{s}_{A}}$ and time taken be ${{t}_{A}}$.
$\Rightarrow {{v}_{A}}=\dfrac{{{s}_{A}}}{{{t}_{A}}}$. ….. (iv).
Here, the magnitude of the aeroplane is 260km. However, the displacement is in the negative direction. Hence, ${{s}_{A}}=-260km$ and ${{t}_{A}}=0.25h$.
Substitute the values in equation (iv).
$\Rightarrow {{v}_{A}}=\dfrac{-260}{0.25}=-1040km{{h}^{-1}}$.
Hence, the average velocity of the aeroplane is -1040km/h.
Note:Note that when a body moves from one point to another, the magnitude of displacement of the body is the shortest distance between the two points. i.e. the length of the line segment joining the two points.
Displacement does not depend on the route or path taken by the body. It only depends on the final and initial positions of the body.
Formula Used:
$u=\dfrac{d}{t}$
$v=\dfrac{s}{t}$
Complete step-by-step answer:
There are two bodies given in the question, the aeroplane and the deluxe bus. The aeroplane travels from Kota to Jaipur and the deluxe bus travels from Jaipur to Kota.
When we travel by air there is no traffic as that of while traveling by road. Therefore, the aeroplane will cover the shortest distance.
It is given that the aeroplane covers a distance of 260 km in 15min and a deluxe bus covers a distance of 320 km in 8 hours.
(i) Let us calculate the speed of the bus.
Average speed of a body is defined as the total distance by that body divided by the total time taken to cover this distance.
Let the average speed of the bus be ${{u}_{B}}$. Let the distance covered by the bus be ${{d}_{B}}$ and time taken be ${{t}_{B}}$.
Hence, the average speed of the bus will be ${{u}_{B}}=\dfrac{{{d}_{B}}}{{{t}_{B}}}$ ….. (i).
It is given that ${{d}_{B}}=320km$ and ${{t}_{B}}=8h$
Substitute the values of ${{d}_{B}}$ and ${{t}_{B}}$ in equations (i).
$\Rightarrow {{u}_{B}}=\dfrac{320}{8}=40km{{h}^{-1}}$
Hence, the average speed of the bus is 40km/h.
(ii) Let us calculate the average velocity of the bus.
Average velocity of a body is defined as the net displacement divided total time taken.
Let the average velocity of the bus be ${{v}_{B}}$. Let the distance covered by the bus be ${{s}_{B}}$ and time taken be ${{t}_{B}}$.
Then the average velocity of the bus is ${{v}_{B}}=\dfrac{{{s}_{B}}}{{{t}_{B}}}$ ….. (ii).
Let us consider the direction from Jaipur to Kota. The magnitude of net displacement of the bus is the length of line segment joining the two cities and it is the shortest distance between the two cities. We know that the aeroplane covers the shortest distance between the two cities. Hence, the displacement of the bus is 320km. ${{s}_{B}}=260km$.
It is given that ${{t}_{B}}=8h$.
Substitute the values in equation (ii).
$\Rightarrow {{v}_{B}}=\dfrac{260}{8}=32.5km{{h}^{-1}}$.
This means that the average velocity of the bus is 32.5km/h.
(iii) Let us calculate the average speed of the aeroplane. We already know what the average speed of a body is.
Let the average speed of the aeroplane be ${{u}_{A}}$. Let the distance covered by the aeroplane be ${{d}_{A}}$ and time taken be ${{t}_{A}}$.
$\Rightarrow {{u}_{A}}=\dfrac{{{d}_{A}}}{{{t}_{A}}}$ …. (iii).
It is given ${{d}_{A}}$=260km and ${{t}_{A}}=15\min =15\times \dfrac{1}{60}h=0.25h$
Substitute the values in equation (iii).
${{u}_{A}}=\dfrac{260}{0.25}=1040km{{h}^{-1}}$.
Hence, the average speed of the aeroplane is 1040km/h.
(iv) Let us calculate the average velocity of the aeroplane.
Let the average velocity of the bus be ${{v}_{A}}$. Let the distance covered by the bus be ${{s}_{A}}$ and time taken be ${{t}_{A}}$.
$\Rightarrow {{v}_{A}}=\dfrac{{{s}_{A}}}{{{t}_{A}}}$. ….. (iv).
Here, the magnitude of the aeroplane is 260km. However, the displacement is in the negative direction. Hence, ${{s}_{A}}=-260km$ and ${{t}_{A}}=0.25h$.
Substitute the values in equation (iv).
$\Rightarrow {{v}_{A}}=\dfrac{-260}{0.25}=-1040km{{h}^{-1}}$.
Hence, the average velocity of the aeroplane is -1040km/h.
Note:Note that when a body moves from one point to another, the magnitude of displacement of the body is the shortest distance between the two points. i.e. the length of the line segment joining the two points.
Displacement does not depend on the route or path taken by the body. It only depends on the final and initial positions of the body.
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