Question

The distance between two places A and B is $180km$ by road and $120km$ by air. An airplane takes $20min$to go from A to B whereas a deluxe bus starting at A takes $5h$ to reach B. Find the average speed and average velocity by both means of transport. (Ignore the vertical ascent of the airplane).

Answer 1

Step 1: Given data

Distance traveled by deluxe bus, $d_r=180km$.

Distance traveled by airplane, $d_a=120km$.

Time taken by airplane to travel between A and B, $t_r=20min=\frac{20}{60}=\frac{1}{3}h$.

Time is taken by deluxe bus to travel between A and B, $t_a=5h$.

The diagrammatic representation of the given data is given below:

Step 2: Calculate the average speed of the bus and airplane

Distance:

1. The complete length of the path between any two points.
2. It is a scalar quantity as it only has magnitude.

Displacement:

1. It is the shortest path between the initial and final positions.
2. It is a vector quantity as it has both direction and magnitude.

The formula for average speed is $\mathit{A}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{s}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\mathbf{=}\frac{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$.

Therefore, the average speed of the bus is:

$\Rightarrow s_r=\frac{180}{5}=36km{h}^{-1}$

Therefore, the average speed of the airplane is:

$\Rightarrow s_a=\frac{120}{\frac{1}{3}}=360km{h}^{-1}$

Step 3: Calculate the average velocity of the bus and airplane

The formula for average velocity is $\mathit{A}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{splacement}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$.

As both the airplane and bus are taking the shortest path to reach point B, their distances will be equal to their displacements.

Therefore, the average velocity of the bus is:

$\Rightarrow v_r=\frac{180}{5}=36km{h}^{-1}$

Therefore, the average velocity of the airplane is:

$\Rightarrow v_a=\frac{120}{\frac{1}{3}}=360km{h}^{-1}$

Thus,

1. The average speed of the bus is $\mathbf{36}\mathbf{}\mathit{k}\mathit{m}{\mathit{h}}^{\mathbf{-}\mathbf{1}}$ and its average velocity is $\mathbf{36}\mathbf{}\mathit{k}\mathit{m}{\mathit{h}}^{\mathbf{-}\mathbf{1}}$.
2. The average speed of the airplane is $\mathbf{360}\mathbf{}\mathit{k}\mathit{m}{\mathit{h}}^{\mathbf{-}\mathbf{1}}$ and its average velocity is $\mathbf{360}\mathbf{}\mathit{k}\mathit{m}{\mathit{h}}^{\mathbf{-}\mathbf{1}}$.