Question

What is Ram’s average speed and average velocity while jogging from X to Z via Y respectively?

$$\begin{array}{rl}& A)2.0{\displaystyle \frac{m}{s}},1.9{\displaystyle \frac{m}{s}}\\ & B)5.1{\displaystyle \frac{m}{s}},9.1{\displaystyle \frac{m}{s}}\\ & C)1.0{\displaystyle \frac{m}{s}},1.0{\displaystyle \frac{m}{s}}\\ & D)1.9{\displaystyle \frac{m}{s}},.0.9{\displaystyle \frac{m}{s}}\end{array}$$

Answer 1

Hint: When a body travels through some distance and displacement then we can first identify the distance and displacement in total time then we will apply the formula for average speed and average velocity. We can get the values of the desired results.
Convert the minutes into seconds.

Complete step-by-step answer:
Average speed is defined as the total distance travelled by the body to the total time taken by the body. It is a scalar quantity which means it is defined by the magnitude only. Its unit is$${\displaystyle \frac{m}{s}}$$.
Average velocity is defined as the total displacement done by the body to the total time taken by the body which means it is defined by the magnitude as well as direction .It is a vector quantity. Its Unit is also same as average speed which is$${\displaystyle \frac{m}{s}}$$.
Let the assume the distance covered by from X to Y be $$d_1$$ and distance covered from Y to Z be$$d_2$$.
Total distance covered by Ram is$$d$$.
Total time taken to cover the distance from point X to Z is$$T$$.
Total Displacement by the Ram from X to Z is$$D$$.
Distance$$d_1$$covered by the Ram from X to Y$$=300m$$
Distance $$d_2$$covered by the Ram from Y to Z $$=100m$$
Total Time taken by the Ram from Point X to Z $$=3min30\sec$$
$$\begin{array}{rl}& T=3min+30\sec \\ & \Rightarrow T=(180+30)\sec s\\ & \therefore T=210\sec s\end{array}$$
Total Distance Covered by the Ram = Distance covered by the Ram from X to Y + Distance covered by the Ram from Y to Z
$$d=d_1+d_2$$
$$\begin{array}{rl}& \Rightarrow d=300+100\\ & \therefore d=400m\end{array}$$
Total Displacement done by the Ram = Distance converted by the Ram from X to Y - Distance Covered by the Ram from Y to Z
$$D=d_1=d_2$$
$$\begin{array}{rl}& \Rightarrow D=300-100\\ & \therefore D=200m\end{array}$$
$$\begin{array}{rl}& Average\mathrm{\_}speed={\displaystyle \frac{Total\mathrm{\_}Dis\tan ce}{Total\mathrm{\_}Time}}\\ & \Rightarrow Average\mathrm{\_}speed={\displaystyle \frac{d}{T}}\\ & \Rightarrow Average\mathrm{\_}speed={\displaystyle \frac{400}{210}}\\ & \therefore Average\mathrm{\_}speed=1.90{\displaystyle \frac{m}{s}}\end{array}$$
$$\begin{array}{rl}& \\ & \Rightarrow Average\mathrm{\_}speed={\displaystyle \frac{Total\mathrm{\_}displacement}{Total\mathrm{\_}time}}\\ & \Rightarrow Average\mathrm{\_}velocity={\displaystyle \frac{D}{T}}\end{array}$$
$$\begin{array}{rl}& \Rightarrow Average\mathrm{\_}velocity={\displaystyle \frac{200}{210}}\\ & \therefore Average\mathrm{\_}velocity=0.9{\displaystyle \frac{m}{s}}\end{array}$$
By the above calculation we can say that the Average speed of the Ram is $$1.9{\displaystyle \frac{m}{s}}$$ and the average Velocity of the ram is$$0.9{\displaystyle \frac{m}{s}}$$.

So, the correct answer is “Option D”.

Note: When average speed is calculated for a very small instant of time then it is called instantaneous speed. This speed is measured by an instrument called a speedometer. This instrument is already installed in our automobiles which are used to get the speed of any instant.