Question

A taxi driver noted reading on the odometer fitted in the vehicle as 1052 km when he started the journey. After a 30 minutes drive, he noted that the odometer reading was 1088 km. What is the average speed of the taxi?
A. $20\,m{s^{ - 1}}$
B. $25\,m{s^{ - 1}}$
C. $30\,m{s^{ - 1}}$
D. $40\,m{s^{ - 1}}$

Answer 1

Use the formula for average velocity of the particle.
Convert the units of distance and time in S.I. unit.
${v_{avg}} = \dfrac{{{x_2} - {x_1}}}{{\Delta t}}$
Here, ${x_2}$ is the final position of the particle, ${x_1}$ is the initial position of the particle and $\Delta t$ is the time taken by the particle for the journey.

Complete step by step answer:
We know, the average speed of the taxi is,
${v_{avg}} = \dfrac{{{x_2} - {x_1}}}{{\Delta t}}$
Here, ${x_2}$ is the final position of the taxi, ${x_1}$ is the initial position of the taxi and $\Delta t$ is the time taken by the taxi for the journey.
Substitute 1088 km for ${x_2}$, 1052 km for ${x_1}$ and 30 min for $\Delta t$ in the above equation.
${v_{avg}} = \dfrac{{1088\,km - 1052\,km}}{{\left( {30\,\min } \right)\left( {\dfrac{{60\,\operatorname{s} }}{{1\,\min }}} \right)}}$
$\Rightarrow {v_{avg}} = \dfrac{{\left( {36\,km} \right)\left( {\dfrac{{1000\,m}}{{1\,km}}} \right)}}{{1800\,\operatorname{s} }}$
$\Rightarrow {v_{avg}} = \dfrac{{36000\,m}}{{1800\,\operatorname{s} }}$
$\therefore {v_{avg}} = 20\,m{s^{ - 1}}$

So, the correct answer is “Option A”.

Additional Information:
The difference between average velocity and instantaneous velocity of the body is the instantaneous velocity is the speed of particle at certain instant of time and the average velocity is the average of velocity in the whole journey of the particle.

Note:
In the given question, the unit of displacement is given in kilometers and the unit of time is given in minutes.
Therefore, convert all the units in S.I. units.