Question

A cyclist travels a distance of 1 km in the first hour, 0.5 km in the second hour and 0.3 km in the third hour. What is the average speed in \[\dfrac{{km}}{{hr}}\] and \[\dfrac{m}{{\sec }}\] ?

Answer 1

Hint: Average speed: The average speed of something is calculated by dividing the total distance travelled by the total time it took to travel that distance.
So, we will apply a mathematical formula for it and keep the given value in the formula.

Complete step by step solution:
Given that,
Distance covered in the first hour is 1 km.
Distance covered in the second hour is 0.5km.
Distance covered in the third hour is 0.3km.
The total distance travelled \[ = 1{\text{ }}km + 0.5{\text{ }}km + 0.3{\text{ }}km = 1.8{\text{ }}km.\]
Total time taken \[ = 1hr + 1hr + 1hr = 3hr\]
Now, \[Average{\text{ }}speed = \dfrac{{Total{\text{ }}distance{\text{ }}traveled}}{{Total{\text{ }}time{\text{ }}taken}}\]
So, keeping value in it,
\[Average{\text{ }}speed = \dfrac{{1.8}}{3}\]
$ = 0.6km{h^{ - 1}}$
Now let's convert \[kmh{r^{ - 1}}\] to \[m{\sec ^{ - 1}}\]
Since,
We know \[1{\text{ }}km = 1000{\text{ }}m\]
and \[1hr = \left( {60 \times 60} \right)s{\text{ }} = 3600s\]
So, \[1kmh{r^{ - 1}} = \dfrac{{1000}}{{60 \times 60}}\]
Now convert above calculated value:
So, \[0.6\dfrac{{km}}{{hr}} = 0.6 \times \dfrac{{1000}}{{3600}}\]
\[ \Rightarrow 0.6 \times \dfrac{{1000}}{{3600}} = \dfrac{{600}}{{3600}}\]
\[ = \dfrac{1}{6}m{s^{ - 1}}\]

Note:
Newton's first law asserts that if a body is at rest or moving in a straight path at a constant speed, it will remain at rest or continue to move in a straight line at a constant speed until acted upon by a force. The law of inertia is the name given to this concept. The second law of Newton is a quantitative description of the effects that a force can have on a body's motion. It asserts that the force imposed on a body equals the time rate of change of its momentum in both magnitude and direction. The product of a body's mass and velocity determines its momentum.