Question

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A man makes three rounds around a square path in 40sec. The length of each side of the square path is 10m. Find average speed and average velocity. Calculate the same thing if he makes 3.5 rounds

Answer 1

Explanations & Calculations

• What you need to know is the definition of the average speed, average velocity & displacement to work this out.

$\qquad\qquad \begin{aligned} \small Average\,Speed&= \small \frac{Total\,Distance\,made}{Total\, time\,spent}\\ \\ \small Average\,Velocity&= \small \frac{Total\,displacement}{Total\,time\,taken}\\ \\ \small Displacement &= \small Shortest\,distance(straight\,indeed)\,between\,2\,points \end{aligned}$

a)

• As he completes 3 rounds around the square, the total distance he covers is

$\qquad\qquad \begin{aligned} \small Total\,distance&= \small 3\times perimeter\,of\,square\\ &= \small 3\times(10\,m\times4)\\ &= \small 3\times40\,\\ &= \small 120\,m \end{aligned}$

• For this he consumes 40 seconds which is the total time.
• Then by the above relationship, we get

$\qquad\qquad \begin{aligned} \small Average\,Speed&= \small \frac{120\,m}{40\,s}\\ &= \small \bold{3\,ms^{-1}} \end{aligned}$

• But when we consider the displacement, he has arrived at the same starting location after travelling even 3 rounds. Which means that his displacement is zero.
• Then by the definition Average Velocity is

$\qquad\qquad \begin{aligned} \small Average\,Velocity&= \small \frac{0\,}{40\,s}=\bold{0\,ms^{-1}} \end{aligned}$

b)

• The total distance he covers at 3.5 rounds is

$\qquad\qquad \begin{aligned} \small Distance&= \small 3.5\times (10\,m\times 4)\\ &= \small 140\,m \end{aligned}$

• Then the average speed is

$\qquad\qquad \begin{aligned} &=\small \frac{140\,m}{40\,s}\\ &= \small \bold{3.5\,ms^{-1}} \end{aligned}$

• But after traveling 3.5 rounds, he stops at the far end of the diagonal of the square drawn from the starting point.
• Then the shortest distance—displacement— is the length of the diagonal of the square, which can be calculated to be (Pythagoras theorem)

$\qquad\qquad \begin{aligned} \small |Diagonal|&= \small\sqrt{(10\,m)^2+(10\,m)^2}\\ &= \small 10\sqrt{2}\,m \end{aligned}$

• Now by the definition, the average velocity can be calculated to be

$\qquad\qquad \begin{aligned} \small Average\,Velocity&= \small \frac{10\sqrt{2}\,m}{40\,s}\\ &= \small \bold{0.354\,ms^{-1}} \end{aligned}$