
It is 260 km from Patna to Ranchi by air and 320 km by road. An airplane takes 30 minutes to go from Patna to Ranchi whereas a deluxe bus takes 8 hours.
(a). Find the average speed of the plane.
(b) Find the average speed of the bus.
(c). Find the average velocity of the plane.
(d). Find the average velocity of the bus.
Answer
583.2k+ views
Hint: As we know that when a vehicle moves, it covers various distances at various time intervals, sometimes the time required to cover the distance is more, and sometimes the time required to cover the distance is less. So to know about the performance of a vehicle we calculate the distance that it covers in per time ratio.
Complete step by step answer:
(a) Given:
(i) The total distance between Patna and Ranchi by air is 260 km.
(ii) The total time taken to cover the distance between Patna and Ranchi is 30 min.
We know that the average speed is defined as:
\[{S_{avg}} = \dfrac{{{D_{total}}}}{{{t_{total}}}}\]…… (I)
Where \[{D_{total}}\] is the total distance travelled and \[{t_{total}}\] is the total time taken.
We can now substitute \[{D_{total}} = 260\,{\text{km}}\] and \[{t_{total}} = 60\,{\text{min}}\] to calculate the average speed of the plane.
\[
\Rightarrow {S_{avg}} = \dfrac{{260\,{\text{km}}}}{{30\,{\text{min}}}} \\
\Rightarrow {S_{avg}} = \dfrac{{260\,{\text{km}}}}{{30\,{\text{min}} \times \dfrac{{\,1\,{\text{hr}}}}{{60\,{\text{min}}}}}} \\
\therefore {S_{avg}} = 520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
\]
$\therefore$ The average speed of the aeroplane is \[520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}}\].
(b) Given:
(i) The distance between Patna and Ranchi by bus is 320 km.
(ii) Time taken to cover the distance between Patna and Ranchi by bus is 8 hrs.
Here we will use equation (i) to determine the average speed of the bus.
\[{S_{avg}} = \dfrac{{{D_{total}}}}{{{t_{total}}}}\] …… (II)
We can now substitute \[{D_{total}} = 320\,{\text{km}}\] and \[{t_{total}} = 8\,{\text{hr}}\] to calculate the average speed of the plane.
\[
\Rightarrow {S_{avg}} = \dfrac{{320\,{\text{km}}}}{{8\;{\text{hr}}}} \\
\therefore {S_{avg}} = 40\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
\]
$\therefore$ The average speed of the bus is \[40\dfrac{{{\text{km}}}}{{{\text{hr}}}}\].
(c) We know that the average velocity is defined by the displacement per unit time. As we know that an airplane travels in a straight path between the source and destination, so its displacement is equal to its distance.
So we can describe the expression for average velocity as:
${V_{avg}} = \dfrac{X}{t}$ …… (III)
Here, ${V_{avg}}$ is the average velocity, $X$ is the displacement and $t$ is the time. We can now substitute $X = 260\,{\text{km}}$ and $t = 30\min = 0.5\,{\text{hr}}$ in equation (III) to find the value of ${V_{avg}}$.
$
{V_{avg}} = \dfrac{{260\,{\text{km}}}}{{0.5\;{\text{hr}}}} \\
\therefore {V_{avg}} = 520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
$
$\therefore$ The average velocity of the plane is $520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}}$.
(d) As we know that to calculate the average velocity of the bus, the displacement between Patna and Ranchi is the air distance i.e. the straight distance which is the displacement equal to 260 km.
Now here, the time taken would remain the same that is equal to 8 hrs
Now we can use equation (III) to find the average velocity and we will substitute $X = 260\,{\text{km}}$ and $t = 8\,{\text{hr}}$ to find the average velocity.
$
\Rightarrow {V_{avg}} = \dfrac{{260\,{\text{km}}}}{{8\,{\text{hr}}}} \\
\therefore {V_{avg}} = 32.5\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
$
$\therefore$ The average velocity of the bus is $32.5\dfrac{{{\text{km}}}}{{{\text{hr}}}}$.
Note:
We have studied in one-dimensional motion that the speed depends on the distance and the velocity depends on the displacement. Speed can never be zero as distance can’t be zero but the velocity can be zero since the displacement can be zero as displacement is the vector quantity and distance is the scalar quantity.
Complete step by step answer:
(a) Given:
(i) The total distance between Patna and Ranchi by air is 260 km.
(ii) The total time taken to cover the distance between Patna and Ranchi is 30 min.
We know that the average speed is defined as:
\[{S_{avg}} = \dfrac{{{D_{total}}}}{{{t_{total}}}}\]…… (I)
Where \[{D_{total}}\] is the total distance travelled and \[{t_{total}}\] is the total time taken.
We can now substitute \[{D_{total}} = 260\,{\text{km}}\] and \[{t_{total}} = 60\,{\text{min}}\] to calculate the average speed of the plane.
\[
\Rightarrow {S_{avg}} = \dfrac{{260\,{\text{km}}}}{{30\,{\text{min}}}} \\
\Rightarrow {S_{avg}} = \dfrac{{260\,{\text{km}}}}{{30\,{\text{min}} \times \dfrac{{\,1\,{\text{hr}}}}{{60\,{\text{min}}}}}} \\
\therefore {S_{avg}} = 520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
\]
$\therefore$ The average speed of the aeroplane is \[520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}}\].
(b) Given:
(i) The distance between Patna and Ranchi by bus is 320 km.
(ii) Time taken to cover the distance between Patna and Ranchi by bus is 8 hrs.
Here we will use equation (i) to determine the average speed of the bus.
\[{S_{avg}} = \dfrac{{{D_{total}}}}{{{t_{total}}}}\] …… (II)
We can now substitute \[{D_{total}} = 320\,{\text{km}}\] and \[{t_{total}} = 8\,{\text{hr}}\] to calculate the average speed of the plane.
\[
\Rightarrow {S_{avg}} = \dfrac{{320\,{\text{km}}}}{{8\;{\text{hr}}}} \\
\therefore {S_{avg}} = 40\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
\]
$\therefore$ The average speed of the bus is \[40\dfrac{{{\text{km}}}}{{{\text{hr}}}}\].
(c) We know that the average velocity is defined by the displacement per unit time. As we know that an airplane travels in a straight path between the source and destination, so its displacement is equal to its distance.
So we can describe the expression for average velocity as:
${V_{avg}} = \dfrac{X}{t}$ …… (III)
Here, ${V_{avg}}$ is the average velocity, $X$ is the displacement and $t$ is the time. We can now substitute $X = 260\,{\text{km}}$ and $t = 30\min = 0.5\,{\text{hr}}$ in equation (III) to find the value of ${V_{avg}}$.
$
{V_{avg}} = \dfrac{{260\,{\text{km}}}}{{0.5\;{\text{hr}}}} \\
\therefore {V_{avg}} = 520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
$
$\therefore$ The average velocity of the plane is $520\,\dfrac{{{\text{km}}}}{{{\text{hr}}}}$.
(d) As we know that to calculate the average velocity of the bus, the displacement between Patna and Ranchi is the air distance i.e. the straight distance which is the displacement equal to 260 km.
Now here, the time taken would remain the same that is equal to 8 hrs
Now we can use equation (III) to find the average velocity and we will substitute $X = 260\,{\text{km}}$ and $t = 8\,{\text{hr}}$ to find the average velocity.
$
\Rightarrow {V_{avg}} = \dfrac{{260\,{\text{km}}}}{{8\,{\text{hr}}}} \\
\therefore {V_{avg}} = 32.5\dfrac{{{\text{km}}}}{{{\text{hr}}}} \\
$
$\therefore$ The average velocity of the bus is $32.5\dfrac{{{\text{km}}}}{{{\text{hr}}}}$.
Note:
We have studied in one-dimensional motion that the speed depends on the distance and the velocity depends on the displacement. Speed can never be zero as distance can’t be zero but the velocity can be zero since the displacement can be zero as displacement is the vector quantity and distance is the scalar quantity.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

