The distance between two points A and B is 100 m. A person moves from A to B with a speed of $20 m s^{- 1}$ and comes back from B to A with a speed of $25 m s^{- 1}$ .
Calculate the average speed and the average velocity of the person. [2 MARKS]

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Solution

Average Speed : 1 Mark
Average Velocity : 1 Mark

Given, distance between A and B, $d = 100 m$
Person moves from A to B with a speed $v_{1} = 20 m s^{- 1}$ and let the time taken be $t_{1}$ .
He moves from B to A with a speed $v_{2} = 25 m s^{- 1}$ and let the time taken be $t_{2}$
Total distance travelled $= 2 d = 2 \times 100 = 200 m$
We know, $s p e e d = \frac{d i s t a n c e}{t i m e} \Rightarrow t i m e = \frac{d i s t a n c e}{s p e e d}$
$∴ t_{1} = \frac{100}{20} = 5 s$ and $t_{2} = \frac{100}{25} = 4 s$
Total time taken, $t = t_{1} + t_{2} = 5 + 4 = 9 s$
We know, $a v e r a g e s p e e d = \frac{t o t a l d i s t a n c e}{t o t a l t i m e} = \frac{200}{9} = 22.2 m s^{- 1}$
$A v e r a g e v e l o c i t y = \frac{t o t a l d i s p l a c e m e n t}{t o t a l t i m e}$
Since the person starts from point A and comes back to point A, total displacement = 0.
Therefore average velocity = 0.

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The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 ms−1 and comes back from B to A with a speed of 25 ms−1. Calculate the average speed and the average velocity of the person. [2 MARKS]

The distance between two points A and B is 100 m. A person moves from A to B with a speed of $20 m s^{- 1}$ and comes back from B to A with a speed of $25 m s^{- 1}$ .
Calculate the average speed and the average velocity of the person. [2 MARKS]