Joseph Jogs from one end $A$ to the other end $B$ of a straight $300 m$ road in $2$ minutes $30$ seconds and then turns around and jogs $100 m$ back to point $C$ in another $1$ minute. What are Joseph's average speeds and velocities in jogging (a) from $A$ to $B$ and (b) from $A$ to $C$ ?

Question

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Answer 1

From pont A to B
Distance covered $= 300 m$
Displacement $= 300 m$
Time taken $= 2 m i n 30 s e c = (2 \times 60) + 30 = 150 s e c$
Average speed $= \frac{D i s t a n c e c o v e r e d}{T i m e t a k e n} = \frac{3 0 0}{1 5 0} = 2 m / s e c$
Average velocity $= \frac{D i s p l a c e m e n t}{T i m e t a k e n} = \frac{3 0 0}{1 5 0} = 2 m / s e c$

From point A to C
Distance covered $= 300 + 100 = 400 m$
Displacement $= 300 - 100 = 200 m$
Time taken $= 3 m i n 30 s e c = (3 \times 60) + 30 = 210 s e c$
Average speed $= \frac{D i s t a n c e c o v e r e d}{T i m e t a k e n} = \frac{4 0 0}{2 1 0} = 1.90 m / s e c$
Average velocity $= \frac{D i s p l a c e m e n t}{T i m e t a k e n} = \frac{2 0 0}{2 1 0} = 0.95 m / s e c$

Answer 2

From pont A to B

Distance covered

= 300 m

Displacement

= 300 m

Time taken

= 2 m i n 30 s e c = (2 \times 60) + 30 = 150 s e c

Average speed

= \frac{D i s t a n c e c o v e r e d}{T i m e t a k e n} = \frac{3 0 0}{1 5 0} = 2 m / s e c

Average velocity

= \frac{D i s p l a c e m e n t}{T i m e t a k e n} = \frac{3 0 0}{1 5 0} = 2 m / s e c

From point A to C

Distance covered

= 300 + 100 = 400 m

Displacement

= 300 - 100 = 200 m

Time taken

= 3 m i n 30 s e c = (3 \times 60) + 30 = 210 s e c

Average speed

= \frac{D i s t a n c e c o v e r e d}{T i m e t a k e n} = \frac{4 0 0}{2 1 0} = 1.90 m / s e c

Average velocity

= \frac{D i s p l a c e m e n t}{T i m e t a k e n} = \frac{2 0 0}{2 1 0} = 0.95 m / s e c