Question

A moving car possesses average velocities $5\,m{s^{ - 1}},10\,m{s^{ - 1}}{\kern 1pt} and{\kern 1pt} 15\,m{s^{ - 1}}$ in the first, second and third seconds respectively. What is the total distance covered by the car in these $3\,s?$
A. $15\,m$
B. $30\,m$
C. $55\,m$
D. None of these

Answer 1

In order to solve this question, we should know that average velocity of a body is not the actual velocity of a body but it’s the average of multiple velocity of a body in given period of time, so here we will calculate the net average velocity of the car in these three seconds and then by using the general relation between velocity, time taken and distance covered, we will figure out the correct option of total distance covered by the moving car.

Formula used:
The relation between velocity distance and time taken is written as
$\text{Distance = velocity} \times \text{time}$

Complete step by step answer:
According to the question, the average velocity in the first second is $5\,m{s^{ - 1}}$. Average velocity in second is $10m{s^{ - 1}}$.Average velocity in third second is $15\,m{s^{ - 1}}$ for these three intervals, net average velocity can be found as taking average of three values which is
$\text{velocity}_{net} = \dfrac{{5 + 10 + 15}}{3}$
$\text{velocity}_{net} = 10\,m{s^{ - 1}}$
Now, we have velocity of $\text{velocity}_{net} = 10\,m{s^{ - 1}}$ and total time taken is $3s$ so, total distance covered is written as
$\text{Distance} = \text{velocity} \times \text{time}$
Putting values we get,
$\text{Distance} = 10 \times 3$
$\therefore \text{Distance} = 30m$
So, the total distance covered by the moving car is $\text{Distance} = 30\,m$.

Hence, the correct option is B.

Note: It should be remembered that, velocity is a vector quantity which has a magnitude as well as direction, but speed is a scalar quantity here, given velocity has only magnitude so we can consider them as their speeds and average of any quantity is the sum of all values and divided by the number of times a sample is taken.