Question

A bus going from Kota to Jaipur passed the $100\,km$ ,$160\,km$ and $220\,km$ points at $10.30\,am$ , $11.30\,am$ and $1.30\,pm$ . Find the average speed of the bus during the interval of $10.30\,am$ and $11.30\,am$ .
A. $60\,Km/h$
B. $30\,Km/h$
C. $60\,Km/h$
D. $60\,Km/h$

Answer 1

Hint- We know that the average speed is calculated as the total distance travelled in the given time interval. In equation form we can write it as
${s_{avg}} = \dfrac{{{\text{total}}\,{\text{distance}}}}{{{\text{time}}\,{\text{taken}}}}$
Using this equation we can find the final answer.

Step by step solution:
In this question we need to find the value of the average speed of the bus in a time interval from $10.30\,am$ to $11.30\,am$ .
It is given that when the bus goes from Kota to Jaipur passes the $100\,km$ points at $10.30\,am$. It passes the $160\,km$ point at $11.30\,am$ and $220\,km$ point at $1.30\,pm$ .
We know that the average speed is calculated as the total distance travelled in the given time interval. Let us denote the average speed as ${s_{avg}}$
In equation form we can write it as
${s_{avg}} = \dfrac{{{\text{total}}\,{\text{distance}}}}{{{\text{time}}\,{\text{taken}}}}$
Since we are asked to find the average speed in the time interval $10.30\,am$ to $11.30\,am$ , we can take the total time as
$t = 11.30 - 10.30 = 1\,h$
Now we need to calculate the total distance travelled in this time. At $10.30\,am$ the bus has crossed the $100\,km$ point and at $11.30\,am$ the bus has crossed the $160\,km$ . Thus the distance travelled in between will be the difference of these two distances.
Thus,
Total distance , $d = 160 - 100 = 60\,km$
Now substitute these values in the equation for finding the average speed .
Then we get,
${s_{avg}} = \dfrac{{60\,km}}{{1\,h}}$
$\therefore {s_{avg}} = 60\,km/h$
This is the value of average speed .

So, the correct answer is option C.

Note: Don’t confuse between average speed and average velocity .Average speed is the ratio of total distance by total time taken. It is a scalar quantity. Whereas, average velocity is the ratio of total displacement to the total time taken. It is a vector quantity. It has both magnitude and direction.